Vectors in Rn form the foundational geometric and algebraic substrate of linear algebra. Ordered n-tuples of real numbers admit two primary operations: vector addition and scalar multiplication. From these elementary operations emerge geometric concepts of length (Euclidean norm), direction, orthogonality, and projections that extend seamlessly into abstract inner product spaces.
Level 1 · Foundations — Vector Space Axioms of Rn, Norms & Normalization
A vector u∈Rn is represented as an n×1 column vector u=u1u2⋮un or an ordered n-tuple (u1,u2,…,un). Addition and scalar multiplication operate componentwise.
Coordinate graph of f(x) = x² − 2 and its instantaneous tangent line at x = 1.
Unit Vector (Normalization): If u=0, the vector u^=∥u∥u satisfies ∥u^∥=1 and points in the direction of u.
Fundamental Axioms of the Euclidean Dot Product
Symmetry (Commutativity)
u⋅v=v⋅u for all u,v∈Rn
Bilinearity (Distributivity)
u⋅(cv+dw)=c(u⋅v)+d(u⋅w)
Positive Definiteness
u⋅u≥0, with u⋅u=0⟺u=0
Scalar Homogeneity of Norm
∥cu∥=∣c∣∥u∥ for any scalar c∈R
Level 1 (Foundations) — Worked Example · Componentwise Calculus & Normalization
Problem: Given vectors u=2−14−2 and v=32−15 in R4, calculate the linear combination w=3u−2v, compute the inner product u⋅v, and find the unit vector u^.
Step 1 · Linear Combination:w=32−14−2−232−15=6−6−3−412−(−2)−6−10=0−714−16.
Step 3 · Euclidean Norm and Normalization:∥u∥=22+(−1)2+42+(−2)2=4+1+16+4=25=5. The normalized unit vector is: u^=51u=2/5−1/54/5−2/5.
Worked Example D — Euclidean Distance & Orthogonal Projection
A displacement has horizontal component Δx=5 and vertical component Δy=5. Find the squared magnitude d2 and the area of the right triangle formed with the axes.
Step-by-step Solution:
By the Pythagorean theorem, d2=(5)2+(5)2=25+25=50.
The enclosed right-triangle area is A=21(5)(5)=12.5.
Level 1 · Foundations
Let a=12−2 and b=k41 in R3. Determine the unique scalar value of k such that a and b are orthogonal (perpendicular).
Two non-zero vectors are orthogonal if and only if their inner product is zero: a⋅b=0. Compute the dot product: (1)(k)+(2)(4)+(−2)(1)=0⟹k+8−2=0⟹k+6=0⟹k=−6.
The Cauchy–Schwarz inequality is the paramount foundational inequality in linear analysis. It guarantees that the geometric definition of the angle between two vectors cosθ=∥u∥∥v∥u⋅v is always mathematically well-defined.
Theorem (Cauchy–Schwarz): For all u,v∈Rn: ∣u⋅v∣≤∥u∥∥v∥, with equality holding if and only if u and v are linearly dependent (one vector is a scalar multiple of the other).
Proof via Quadratic Discriminant: If v=0, then u⋅0=0 and ∥u∥∥0∥=0, so equality holds trivially. Assume v=0. For any real parameter t∈R, consider the vector w(t)=u−tv. By the positive definiteness of the Euclidean norm: 0≤∥u−tv∥2=(u−tv)⋅(u−tv). Expanding by bilinearity and symmetry: 0≤u⋅u−2t(u⋅v)+t2(v⋅v)=∥v∥2t2−2(u⋅v)t+∥u∥2. This is a quadratic polynomial p(t)=At2+Bt+C in t, where A=∥v∥2>0, B=−2(u⋅v), and C=∥u∥2. Because p(t)≥0 for all t∈R, the parabola cannot cross the t-axis twice, which requires that its discriminant Δ=B2−4AC be non-positive: Δ=[−2(u⋅v)]2−4(∥v∥2)(∥u∥2)≤0.4(u⋅v)2−4∥u∥2∥v∥2≤0⟹(u⋅v)2≤∥u∥2∥v∥2. Taking the positive square root of both sides establishes: ∣u⋅v∣≤∥u∥∥v∥.■
● Corollary 1.2 · The Triangle Inequality in Rn
Theorem (Minkowski's Triangle Inequality): For all u,v∈Rn: ∥u+v∥≤∥u∥+∥v∥.
Proof: Compute the squared norm using inner products: ∥u+v∥2=(u+v)⋅(u+v)=∥u∥2+2(u⋅v)+∥v∥2. Applying the Cauchy–Schwarz inequality, u⋅v≤∣u⋅v∣≤∥u∥∥v∥: ∥u+v∥2≤∥u∥2+2∥u∥∥v∥+∥v∥2=(∥u∥+∥v∥)2. Taking the non-negative square root yields ∥u+v∥≤∥u∥+∥v∥. ■
⊙ Explorer · Dot product & Cauchy–Schwarzinteractive
u, v, their dot product, and the Cauchy-Schwarz bound |u||v|
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u · v4.023
|u| |v| (bound)5.000
u = ⟨1, 2⟩ is fixed; v keeps the same length as Worked example 1.B's v = ⟨2, 1⟩ (|v| = √5) but rotates freely. Drag near θ ≈ 63.43° and observe u·v approach the bound |u||v| = 5, the equality case in Cauchy-Schwarz where the vectors are parallel.
Orthogonal vectors u=⟨3,2⟩ and v=⟨−2,3⟩ in R2. Their dot product u⋅v=3(−2)+2(3)=0 confirms that the angle between them is 90∘ and their directions are perpendicular.
Level 2 (Medium) — Worked Example · Orthogonal Projection & Vector Decomposition
Problem: Decompose u=512 into the sum of two orthogonal vectors u=u∥+u⊥, where u∥ is parallel to v=12−1 and u⊥ is orthogonal to v.
Level 3 · Advanced / University Rigor — Inner Product Identities, Parallelogram Law & Hyperplanes
In advanced linear algebra and functional analysis, an inner product space (V,⟨⋅,⋅⟩) is characterized by geometric identities that relate the induced norm to the underlying inner product. Chief among these are the Parallelogram Law and the Polarization Identity.
● Theorem 1.3 & Formal Analytical Proof · The Parallelogram Law & Polarization Identity
Theorem (Parallelogram Law): In any real inner product space, the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals: ∥u+v∥2+∥u−v∥2=2∥u∥2+2∥v∥2.
Proof: Expanding both squared norms by inner product bilinearity: ∥u+v∥2=∥u∥2+2⟨u,v⟩+∥v∥2,∥u−v∥2=∥u∥2−2⟨u,v⟩+∥v∥2. Adding the two equations eliminates the cross-term 2⟨u,v⟩: ∥u+v∥2+∥u−v∥2=2∥u∥2+2∥v∥2.■
Theorem (Polarization Identity): The inner product can be completely recovered from the norm function: ⟨u,v⟩=41(∥u+v∥2−∥u−v∥2).
Significance: By the Jordan–von Neumann Theorem, a normed vector space (V,∥⋅∥) can be equipped with an inner product that induces the norm if and only if the norm satisfies the Parallelogram Law.
⚠️ Common Traps & University Pitfalls · Vectors in Rⁿ
Canceling Dot Products:u⋅v=u⋅w does NOT imply v=w. It only implies u⋅(v−w)=0, meaning u is orthogonal to v−w. Dot product division is mathematically undefined!
Norm of a Sum:∥u+v∥2=∥u∥2+∥v∥2 in general. The Pythagorean Theorem ∥u+v∥2=∥u∥2+∥v∥2 holds if and only if u⋅v=0 (orthogonality).
Associativity Fallacy: The expression u⋅v⋅w is syntax nonsense because u⋅v is a scalar; you cannot take the dot product of a scalar with the vector w. The vector scalar multiplication is (u⋅v)w.
Level 3 (Advanced) — Worked Example · Hyperplane Equations & Point-to-Plane Distance in Rⁿ
Problem: In R4, a hyperplane H is defined by the equation 2x1−x2+3x3+2x4=14. 1. Identify the normal vector n and express H in vector form n⋅(x−p)=0. 2. Prove analytically and calculate the shortest Euclidean distance from the point x0=(1,2,0,1) to H.
Step 1 · Normal Vector and Plane Representation: The coefficients define the normal vector n=2−132. For any point p∈H, n⋅p=14. For instance, choosing p=(7,0,0,0) gives n⋅p=14. Thus H={x∈R4:n⋅(x−p)=0}.
Step 2 · Analytical Distance Derivation: The vector connecting any p∈H to x0 is x0−p. The distance to the hyperplane is the length of the projection of x0−p onto the unit normal n^: d(x0,H)=∥n∥∣n⋅(x0−p)∣=∥n∥∣n⋅x0−n⋅p∣=∥n∥∣n⋅x0−14∣.
Let u,v∈Rn satisfy ∥u∥=4, ∥v∥=5, and ∥u+v∥=7. Using the Parallelogram Law, find the exact length of the difference vector ∥u−v∥, and determine the angle θ between u and v.
Step 1 · Apply the Parallelogram Law:∥u+v∥2+∥u−v∥2=2∥u∥2+2∥v∥2. Substitute known values: 72+∥u−v∥2=2(42)+2(52)⟹49+∥u−v∥2=2(16)+2(25)=32+50=82.∥u−v∥2=82−49=33⟹∥u−v∥=33.
Step 2 · Determine the angle θ: Using the Polarization Identity: u⋅v=41(∥u+v∥2−∥u−v∥2)=41(49−33)=416=4. Now compute cosθ: cosθ=∥u∥∥v∥u⋅v=(4)(5)4=51=0.2⟹θ=arccos(0.2)≈78.46∘.
[Vectors in Rn & Inner Product Spaces · Practice 4] If the governing relation for this system is g(t)=6t+6, evaluate g(10) and find the inverse input t when g(t)=72.
📌 Linear Algebra · Chapter 2Core Technique · Augmented matrices & elementary row operations · REF vs. RREF & Gauss–Jordan algorithm · Pivot existence theorem & parametric vector solutions
Systems of linear equations represent the most widely used mathematical construct in quantitative science. Any linear system can be encoded compactly as an augmented matrix [A∣b]. Through Gaussian and Gauss–Jordan elimination, elementary row operations transform the system into echelon form while preserving its solution set, revealing existence, uniqueness, and parametric solution spaces.
A system of m linear equations in n unknowns x1,x2,…,xn can be written as: a11x1+a12x2+⋯+a1nxna21x1+a22x2+⋯+a2nxnam1x1+am2x2+⋯+amnxn=b1=b2⋮=bm⟺Ax=b.
Coordinate graph of f(x) = x² − 2 and its instantaneous tangent line at x = 1.
● Definition 2.1 — Augmented Matrix, REF & RREF
1. Augmented Matrix: The m×(n+1) rectangular array: [A∣b]=a11a21⋮am1a12a22⋮am2……⋱…a1na2n⋮amn∣∣∣∣b1b2⋮bm.
2. Row Echelon Form (REF): A matrix is in REF if: • All all-zero rows are grouped at the very bottom. • Each leading non-zero entry (called a pivot) of a non-zero row is strictly to the right of the pivot in the row above it. • All entries in a column directly below a pivot are zero.
3. Reduced Row Echelon Form (RREF): A matrix is in RREF if it satisfies REF plus: • Every pivot entry is equal to 1. • Each pivot 1 is the only non-zero entry in its entire column (entries above and below are zero).
The Three Elementary Row Operations
Row Interchange (Swap)
Ri⟷Rj: Swap equation i and equation j
Row Scaling
Ri⟵cRi with c=0: Multiply equation i by a non-zero scalar
Row Replacement (Elimination)
Ri⟵Ri+cRj: Add a scalar multiple of row j to row i
Level 1 (Foundations) — Worked Example · Gauss–Jordan Elimination to RREF
Problem: Solve the linear system using Gauss–Jordan elimination: x1+2x2−x32x1+5x2+x33x1+7x2+2x3=1=7=12
Step 1 · Form the Augmented Matrix:[A∣b]=123257−112∣∣∣1712.
Conclusion: The system has the unique solution x1=5,x2=−1,x3=2.
Worked Example D — Euclidean Distance & Orthogonal Projection
A displacement has horizontal component Δx=5 and vertical component Δy=5. Find the squared magnitude d2 and the area of the right triangle formed with the axes.
Step-by-step Solution:
By the Pythagorean theorem, d2=(5)2+(5)2=25+25=50.
The enclosed right-triangle area is A=21(5)(5)=12.5.
Level 1 · Foundations
Convert the matrix into Reduced Row Echelon Form (RREF): M=(12−2−415).
Level 2 · Medium — The Existence & Uniqueness Theorem and Solution Invariance
Every linear system has either no solution, exactly one unique solution, or infinitely many solutions. This fundamental trichotomy is determined completely by the pivot positions in the augmented matrix.
● Theorem 2.1 & Proof · Invariance of Solution Sets Under Row Operations
Theorem: Applying any sequence of elementary row operations to the augmented matrix [A∣b] preserves the solution set of the linear system Ax=b.
Proof: It suffices to verify each elementary row operation individually: • Row Interchange (Ri↔Rj): Swapping two equations changes only their presentation order in the conjunction of statements E1∧⋯∧Em. Conjunction is commutative, so the solution set is unchanged. • Row Scaling (Ri←cRi,c=0): If x satisfies equation Ei:∑aikxk=bi, multiplying both sides by c shows x satisfies cEi. Conversely, since c=0, multiplying cEi by 1/c recovers Ei. Hence x satisfies Ei⟺x satisfies cEi. • Row Replacement (Ri←Ri+cRj,j=i): If x satisfies both Ei and Ej, it satisfies Ei+cEj. Conversely, if x satisfies the new system, it satisfies Ej (which was untouched) and Ei+cEj; subtracting cEj recovers the original equation Ei. Since each operation is reversible and preserves the solution set, any finite composition of row operations preserves the exact solution set. ■
● Theorem 2.2 · Existence and Uniqueness (Rouché–Capelli Theorem)
Let [A∣b] be the augmented matrix of a linear system with n unknowns, and let [U∣c] be its row echelon form.
Consistency Criterion: A linear system is consistent (possesses at least one solution) if and only if the rightmost column of the augmented matrix is NOT a pivot column. In other words, there is no row of the form: (00…0∣d)with d=0.
Uniqueness vs. Infinitude: If the system is consistent: • Unique Solution: Exactly when every column of A contains a pivot (no free variables, rank(A)=n). • Infinitely Many Solutions: When there is at least one column of A without a pivot (at least one free variable, rank(A)<n).
⊙ Explorer · Two lines, one systeminteractive
The two equations, the system's type, and its solution
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system typeunique solution
solution(3.000, 2.000)
Eq. 1 is fixed at x + y = 5 (indigo). Eq. 2 is a·x − y = c (amber), which at default a = 1, c = 1 is x − y = 1. Drag a to −1 to make the lines parallel, then adjust c to observe both dependent (c = −5, identical lines) and inconsistent (c ≠ −5, parallel lines) cases.
Geometric representation of the linear system x+y=5 and x−y=1. The unique solution (3,2) is the single point where the two lines intersect.
Level 2 (Medium) — Worked Example · Parametric Vector Form for Consistent Underdetermined Systems
Problem: Express the solution set of the system in parametric vector form: x1−2x2+x3+3x42x1−4x2+3x3+8x4=4=11
Step 1 · Reduce Augmented Matrix to RREF:(12−2−41338∣∣411)R2←R2−2R1(10−201132∣∣43). Eliminate above pivot 2 via R1←R1−R2: (10−200112∣∣13).
Step 2 · Identify Basic and Free Variables: Pivots are in columns 1 and 3 ⟹x1,x3 are basic variables. Columns 2 and 4 lack pivots ⟹x2=s and x4=t are free parameters (s,t∈R).
Step 3 · Express Basic Variables in terms of Free Variables:x1=1+2s−t,x3=3−2t.
Geometric Interpretation: The solution set is a 2-dimensional affine plane in R4 passing through particular solution xp=(1,0,3,0)T spanned by the null space vectors.
Level 2 · Medium
Find all solutions to the homogeneous linear system: x1+3x2−2x32x1+6x2−4x3=0=0
Step 1: Set up and row reduce the coefficient matrix: (1236−2−4)R2←R2−2R1(1030−20).
Step 2: Pivot is in column 1 (x1 basic); columns 2 and 3 have no pivots (x2=s,x3=t free).
Step 3: Solve for x1=−3s+2t. Write in parametric vector form: x=x1x2x3=s−310+t201,s,t∈R.
In advanced applications, linear systems frequently contain physical or engineering parameters k∈R. Determining bifurcation points where a system transitions between unique, infinite, and zero solutions requires rigorous analysis of leading variables and determinant degeneration.
● Theorem 2.3 · Affine Structure of Solution Sets
Theorem: Let Ax=b be a consistent linear system, and let xp be any particular solution such that Axp=b. Then the complete solution set S is given by: S=xp+Null(A)={xp+xh:Axh=0}.
Proof: 1. Suppose x=xp+xh with Axh=0. Then: Ax=A(xp+xh)=Axp+Axh=b+0=b. Thus every vector of the form xp+xh is a solution. 2. Conversely, let x be any arbitrary solution satisfying Ax=b. Define xh=x−xp. Then: Axh=A(x−xp)=Ax−Axp=b−b=0. Thus xh∈Null(A), so x=xp+xh∈xp+Null(A). ■
⚠️ Common Traps & University Pitfalls · Linear Systems
Dividing by Parameter Expressions: During row operations with parameter k, performing Ri←k−21Ri assumes k=2. You must explicitly bifurcate your analysis: treat k=2 as an independent case before dividing!
Overdetermined vs. Inconsistent: Having more equations than unknowns (m>n) does NOT mean the system has no solution. If equations are linearly dependent, an overdetermined system can have a unique solution or even infinitely many solutions.
Free Variables vs. Free Parameters: A variable is free only if its corresponding column in A lacks a pivot in REF. Never arbitrarily choose a basic variable as a free parameter.
Level 3 (Advanced) — Worked Example · Complete Parameter Bifurcation Analysis
Problem: Determine all values of the parameter k∈R for which the linear system has (a) a unique solution, (b) infinitely many solutions, or (c) no solution: x+y+kz3x+4y+2z2x+3y−z=2=k=1
Step 1 · Set up Augmented Matrix and Row Reduce:132143k2−1∣∣∣2k1R2←R2−3R1R3←R3−2R1100111k2−3k−1−2k∣∣∣2k−6−3.
Step 3 · Analyze Row 3: (k−3)z=3−k: • Case 1 (k=3): The coefficient k−3=0, so there is a pivot in row 3 column 3. All three columns of A have pivots ⟹Unique solution for all k=3 (with z=−1). • Case 2 (k=3): Row 3 becomes: (003−3∣3−3)=(000∣0). The system is consistent and column 3 lacks a pivot (z is free) ⟹Infinitely many solutions when k=3. • Case 3 (No Solution): For row 3 to represent a contradiction 0=d=0, we would require k−3=0 and 3−k=0, which is impossible since 3−3=0. Hence there is no value of k for which the system is inconsistent.
Level 3 · Advanced / University Rigor
Find the general solution in parametric vector form for the underdetermined linear system: x1+2x2−3x3+x4+2x5−x1−2x2+4x3−2x4−x52x1+4x2−5x3+x4+5x5=3=−2=7
Step 1 · Set up Augmented Matrix and Eliminate:1−122−24−34−51−212−15∣∣∣3−27R2←R2+R1R3←R3−2R1100200−3111−1−1211∣∣∣311. Perform R3←R3−R2 and R1←R1+3R2 to reach RREF: 100200010−2−10510∣∣∣610.
Step 2 · Identify Basic and Free Variables: Pivots in columns 1 and 3 ⟹x1,x3 are basic. Columns 2, 4, 5 lack pivots ⟹x2=r,x4=s,x5=t are free parameters.
[Systems of Linear Equations & Row Reduction · Practice 4] If the governing relation for this system is g(t)=6t+6, evaluate g(10) and find the inverse input t when g(t)=72.
A matrix is both an algebraic data structure for linear equations and the concrete representation of a linear map between vector spaces. Matrix multiplication embodies the composition of linear mappings, while matrix inversion represents the exact reversal of that transformation. This chapter develops the complete algebra of matrices, elementary matrix factorizations, and the unifying Invertible Matrix Theorem.
An m×n matrix A consists of m rows and n columns. It can be viewed horizontally as a stack of row vectors or vertically as an array of column vectors: A=[a1a2…an].
Coordinate graph of f(x) = x² − 2 and its instantaneous tangent line at x = 1.
● Definition 3.1 — Matrix Operations & The Two Perspectives of Matrix Multiplication
1. Entrywise Operations: For matrices A,B∈Rm×n and c∈R: • Addition: (A+B)ij=aij+bij. • Scalar Multiplication: (cA)ij=caij.
2. Matrix-Vector Product (Column Combination Perspective): For x=(x1,…,xn)T∈Rn: Ax=x1a1+x2a2+⋯+xnan. The vector Ax is a linear combination of the columns of A with weights from x.
3. Matrix Multiplication: For A∈Rm×n and B∈Rn×p: • Dot Product Rule: The (i,k)-th entry of AB is: (AB)ik=j=1∑naijbjk=Rowi(A)⋅Colk(B). • Column Perspective:AB=[Ab1Ab2…Abp]. Each column of AB is A applied to a column of B.
● Theorem 3.1 & Analytical Proof · Transpose of a Matrix Product
Theorem: For any A∈Rm×n and B∈Rn×p: (AB)T=BTAT. (Notice the order of multiplication is strictly reversed).
Proof: Let C=AB. The (i,k)-th entry of CT=(AB)T is the (k,i)-th entry of C: (CT)ik=Cki=(AB)ki=j=1∑nAkjBji. Now consider the product D=BTAT. The (i,k)-th entry of D is: Dik=j=1∑n(BT)ij(AT)jk=j=1∑nBjiAkj=j=1∑nAkjBji. Since (CT)ik=Dik for all i,k, it follows that (AB)T=BTAT. ■
Level 1 (Foundations) — Worked Example · Matrix Multiplication & Non-Commutativity
Problem: Let A=(1324) and B=(02−11). Compute AB and BA, verify that AB=BA, and verify that (AB)T=BTAT.
Worked Example D — Euclidean Distance & Orthogonal Projection
A displacement has horizontal component Δx=5 and vertical component Δy=5. Find the squared magnitude d2 and the area of the right triangle formed with the axes.
Step-by-step Solution:
By the Pythagorean theorem, d2=(5)2+(5)2=25+25=50.
The enclosed right-triangle area is A=21(5)(5)=12.5.
Level 1 · Foundations
Let A=(24−103−2) and x=3−21. Compute Ax explicitly using the linear combination of columns definition.
Using the column combination perspective: Ax=3(24)+(−2)(−10)+1(3−2)=(612)+(20)+(3−2)=(1110).
Level 2 · Medium — Matrix Inverses, Invertibility of 2×2 Matrices & The Gauss–Jordan Algorithm
A square n×n matrix A is invertible (non-singular) if it represents a bijection on Rn. Inverting A allows direct algebraic solution of linear systems: Ax=b⟺x=A−1b.
● Theorem 3.2 & Proof · Inversion of 2×2 Matrices & The Product Inversion Rule
Theorem 1 (2×2 Inversion): The matrix A=(acbd) is invertible if and only if det(A)=ad−bc=0. Its inverse is: A−1=ad−bc1(d−c−ba).
Proof: Let Δ=ad−bc=0. Multiply A by B=Δ1(d−c−ba): AB=Δ1(acbd)(d−c−ba)=Δ1(ad−bccd−dc−ab+ba−cb+da)=Δ1(Δ00Δ)=(1001)=I. Similarly, computing BA gives I. Thus B=A−1. ■
Theorem 2 (Inverse of a Product): If A and B are invertible n×n matrices, then AB is invertible and: (AB)−1=B−1A−1.
Proof: Using associativity: (AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I.(B−1A−1)(AB)=B−1(A−1A)B=B−1IB=B−1B=I. By uniqueness of inverses, (AB)−1=B−1A−1. ■
⊙ Explorer · Matrix transformation sandboxinteractive
The matrix, its column images, determinant, and orientation
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det(A)1.000
area scale factor1.000
The indigo arrow is A's first column A e₁; the amber arrow is its second column A e₂. The filled parallelogram represents the image of the unit square, with area equal to |det(A)|. When det(A) < 0, the fill switches to amber, reflecting orientation reversal.
Action of matrix A=(3112) on standard basis vectors e1=⟨1,0⟩ and e2=⟨0,1⟩. The columns of A define the transformed coordinate axes.
Level 2 (Medium) — Worked Example · Matrix Inversion via Gauss–Jordan [A | I] Algorithm
Problem: Invert the 3×3 matrix A=1240−11238 using row reduction on [A∣I].
Step 1 · Augment with Identity Matrix:[A∣I]=1240−11238∣∣∣100010001.
Step 2 · Forward Elimination: Perform R2←R2−2R1 and R3←R3−4R1: 1000−112−10∣∣∣1−2−4010001. Scale R2←−R2: 100011210∣∣∣12−40−10001. Eliminate in row 3: R3←R3−R2: 10001021−1∣∣∣12−60−11001R3←−R3100010211∣∣∣1260−1−100−1.
Step 3 · Backward Elimination to RREF: Eliminate column 3 above row 3: R2←R2−R3 and R1←R1−2R3: 100010001∣∣∣1−2(6)2−660−2(−1)−1−(−1)−10−2(−1)0−(−1)−1=100010001∣∣∣−11−4620−121−1.
Conclusion:A−1=−11−4620−121−1.
Level 2 · Medium
Let A=(3152) and b=(41). Compute A−1 and use it to solve the linear system Ax=b.
Level 3 · Advanced / University Rigor — Elementary Matrices & The Invertible Matrix Theorem
Every elementary row operation corresponds to pre-multiplying by an elementary matrix E. This establishes the profound equivalence between row reduction algorithms, matrix factorizations, and the structural properties of linear mappings.
● Theorem 3.3 · Elementary Matrix Decomposition
An elementary matrixE is obtained by performing a single elementary row operation on the identity matrix I. Pre-multiplying any matrix A by E executes that exact row operation on A: RowOp(A)=EA.
Every elementary matrix E is invertible, and its inverse E−1 is the elementary matrix of the same type that reverses the operation. Consequently, an n×n matrix A is invertible if and only if it is row equivalent to In, in which case A can be factored as a product of elementary matrices: A=E1E2…Ek.
● Theorem 3.4 · The Invertible Matrix Theorem (IMT)
Let A be a square n×n matrix. The following statements are mathematically equivalent (either all are true, or all are false):
A is an invertible matrix.
A is row equivalent to the n×n identity matrix In.
A has n pivot positions (rank A=n).
The equation Ax=0 has only the trivial solution x=0.
The columns of A form a linearly independent set.
The linear transformation T(x)=Ax is one-to-one (injective).
The equation Ax=b has at least one solution for each b∈Rn.
The columns of A span Rn.
The linear transformation T(x)=Ax maps Rn onto Rn (surjective).
There is an n×n matrix C such that CA=I.
There is an n×n matrix D such that AD=I.
AT is an invertible matrix.
det(A)=0.
The number 0 is not an eigenvalue of A.
Col(A)=Rn and Null(A)={0}.
⚠️ Common Traps & University Pitfalls · Matrices
Assuming Zero Product Property: If AB=0, it does NOT imply A=0 or B=0. For example, (0010)(0010)=(0000). Non-zero matrices can have zero products!
Cancellation Fallacy: If AB=AC and A=0, you CANNOT cancel A to conclude B=C. Cancellation is legal only if A is invertible (det(A)=0).
Order of Inverses and Transposes: Remembering socks and shoes: (AB)−1=B−1A−1 and (AB)T=BTAT. Writing (AB)−1=A−1B−1 is a fatal error in non-commutative matrix algebra.
Level 3 (Advanced) — Worked Example · One-Sided Inverses for Square Matrices
Problem: Let A and B be n×n square matrices such that AB=I. Prove rigorously that BA=I and therefore B=A−1.
Step 1 · Consider the Homogeneous Equation Bx=0: Suppose Bx=0 for some vector x∈Rn. Pre-multiply by A: A(Bx)=A0⟹(AB)x=0⟹Ix=0⟹x=0. Thus Bx=0 has only the trivial solution.
Step 2 · Apply the Invertible Matrix Theorem: Since B is an n×n square matrix whose homogeneous equation has only the trivial solution, by the IMT (statement 4 ⟹ statement 1), B is invertible, so B−1 exists.
Step 3 · Compute BA: Multiply AB=I on the left by B−1 and on the right by B: B−1(AB)B=B−1(I)B⟹(B−1AB)B=B−1B=I. More directly, multiply AB=I on the left by B and on the right by B−1: B(AB)B−1=B(I)B−1⟹(BA)(BB−1)=BB−1⟹(BA)I=I⟹BA=I. Thus B is the genuine two-sided inverse of A. ■
Level 3 · Advanced / University Rigor
Suppose an n×n matrix A satisfies the polynomial equation A3−4A2+2A−5I=0. Prove that A is invertible, and express A−1 as an explicit polynomial in A.
Step 1 · Isolate the Identity Matrix:A3−4A2+2A=5I. Divide both sides by 5: 51(A3−4A2+2A)=I.
Step 2 · Factor out A:A[51(A2−4A+2I)]=I. Similarly, since polynomials in A commute: [51(A2−4A+2I)]A=I.
Conclusion: By the definition of matrix inverse, A is invertible, and its inverse is: A−1=51A2−54A+52I.
[Matrices, Matrix Algebra & Inverses · Practice 4] If the governing relation for this system is g(t)=6t+6, evaluate g(10) and find the inverse input t when g(t)=72.
The determinant is the unique multilinear, alternating scalar functional on square matrices that measures the volume scaling factor and orientation change induced by a linear transformation. Its non-zero status completely characterizes matrix invertibility, while its algebraic properties provide the foundational connection between algebra and geometry.
The determinant can be defined axiomatically through three characterizing properties, or constructively via recursive cofactor expansion along any row or column.
Coordinate graph of f(x) = x² − 2 and its instantaneous tangent line at x = 1.
1. The Three Axioms of the Determinant: The determinant det:Rn×n→R is the unique function satisfying: • Multilinearity in Rows:det is linear with respect to each individual row when all other rows are held fixed. • Alternating Property: Swapping any two rows negates the determinant: det(R1,…,Rj,…,Ri,…,Rn)=−det(R1,…,Ri,…,Rj,…,Rn). If two rows are identical, det(A)=0. • Normalization:det(In)=1.
2. Laplace Cofactor Expansion: For any chosen row i (or column j): det(A)=j=1∑naijCij=j=1∑n(−1)i+jaijdet(Mij), where Mij is the (n−1)×(n−1) minor submatrix formed by deleting row i and column j, and Cij=(−1)i+jdet(Mij) is the cofactor.
3. Triangular Matrices: If A is upper-triangular, lower-triangular, or diagonal, its determinant is simply the product of its main diagonal entries: det(A)=a11a22…ann=i=1∏naii.
Level 1 (Foundations) — Worked Example · Cofactor Expansion Strategy
Problem: Evaluate det(A) for the 3×3 matrix by expanding along the row or column with the most zeros: A=20154−2−306.
Step 1 · Choose Optimal Expansion Path: Row 2 contains two zeros (a21=0,a23=0), leaving only a22=4.
Step 2 · Apply Laplace Expansion along Row 2:det(A)=(−1)2+1(0)det(M21)+(−1)2+2(4)det(M22)+(−1)2+3(0)det(M23).det(A)=4det(M22)=421−36.
Worked Example D — Euclidean Distance & Orthogonal Projection
A displacement has horizontal component Δx=5 and vertical component Δy=5. Find the squared magnitude d2 and the area of the right triangle formed with the axes.
Step-by-step Solution:
By the Pythagorean theorem, d2=(5)2+(5)2=25+25=50.
The enclosed right-triangle area is A=21(5)(5)=12.5.
Level 1 · Foundations
Evaluate the determinant of the 4×4 triangular matrix: T=3000−2−100752048−35.
Since T is upper-triangular, its determinant is the product of its diagonal entries: det(T)=(3)(−1)(2)(5)=−30.
Level 2 · Medium — Geometric Volume Interpretation & The Multiplicative Property
Geometrically, ∣det(A)∣ represents the hypervolume of the n-dimensional parallelotope spanned by the column vectors of A. Under a linear mapping T(x)=Ax, any region S⊂Rn with volume V(S) is mapped to an image with volume ∣det(A)∣V(S).
⊙ Explorer · Determinant as signed areainteractive
u, v, their determinant, the parallelogram's area, and invertibility
quantity
value
Drag to rotate, scroll to zoom · loads a small 3D library on first use
det([u v])5.000
area = |det|5.000
matrix [u v]invertible
u and v are the columns of the matrix [u v]; det([u v]) = uₓvy − uyvx is the signed area. Drag u onto v and the parallelogram collapses to a line of area 0, showing that the matrix is singular when its columns are linearly dependent.
Parallelogram in R2 spanned by vectors u=⟨3,0⟩ and v=⟨1,2⟩. The determinant det(3012)=(3)(2)−(1)(0)=6 represents the exact geometric area of the parallelogram.
Theorem: For any n×n matrices A and B: det(AB)=det(A)det(B).
Proof via Elementary Matrices: 1. If E is an elementary matrix: • Row interchange: det(E)=−1, and det(EA)=−det(A)=det(E)det(A). • Row scaling by c: det(E)=c, and det(EA)=cdet(A)=det(E)det(A). • Row replacement: det(E)=1, and det(EA)=det(A)=det(E)det(A). Hence for any elementary matrix E, det(EB)=det(E)det(B). 2. By induction, if A=E1E2…Ek is invertible: det(AB)=det(E1…EkB)=det(E1)…det(Ek)det(B)=det(E1…Ek)det(B)=det(A)det(B). 3. If A is not invertible, then rank(A)<n⟹rank(AB)≤rank(A)<n, so AB is also not invertible. Thus det(AB)=0 and det(A)det(B)=0⋅det(B)=0. Therefore det(AB)=det(A)det(B) holds universally. ■
Corollaries: • Transpose: det(AT)=det(A). • Inverse: If A is invertible, det(A−1)=det(A)1. • Scalar Multiple: det(cA)=cndet(A) for n×n matrix A.
Level 2 (Medium) — Worked Example · Fast Determinant via Row Reduction
Problem: Evaluate the determinant of A=12012432−11223514 using Gaussian row operations.
Step 1 · Row Replacement on Column 1: Perform R2←R2−2R1 and R4←R4−R1 (row replacements leave determinant unchanged): det(A)=10002030−13233−111.
Step 2 · Row Swap to place pivot in Row 2: Perform R2↔R3 (multiplies determinant by −1): det(A)=−10002300−123331−11.
Step 3 · Eliminate Column 3 in Row 4: Perform R4←R4−R3 (no change to determinant): det(A)=−10002300−123031−12.
Let A and B be 3×3 matrices with det(A)=3 and det(B)=−2. Compute the exact numerical value of: det(2A2B−1AT).
Apply determinant algebraic properties for 3×3 matrices: det(2M)=23det(M)=8det(M). Using the multiplicative, power, transpose, and inverse properties: det(A2B−1AT)=[det(A)]2⋅det(B)1⋅det(A)=det(B)[det(A)]3. Substitute det(A)=3 and det(B)=−2: det(2A2B−1AT)=8⋅−233=8⋅−227=8(−13.5)=−108.
Level 3 · Advanced / University Rigor — The Classical Adjugate, Cramer's Rule & Invertibility Invariant
The adjugate matrix provides an explicit analytical formula for the inverse of an arbitrary n×n matrix, leading to Cramer's Rule for solving non-singular linear systems without row operations.
● Theorem 4.2 & Proof · The Fundamental Adjugate Identity & Cramer's Rule
Definition: Let C be the cofactor matrix of A with entries Cij=(−1)i+jdet(Mij). The adjugate matrixadj(A) is the transpose of the cofactor matrix: adj(A)=CT,(adj(A))ij=Cji.
Theorem (Fundamental Adjugate Identity): For any square matrix A∈Rn×n: Aadj(A)=adj(A)A=det(A)In. Consequently, if det(A)=0, then A is invertible and: A−1=det(A)1adj(A).
Proof: Consider the (i,k)-th entry of the product Aadj(A): (Aadj(A))ik=j=1∑naij(adj(A))jk=j=1∑naijCkj. • If i=k, this is precisely the Laplace cofactor expansion of det(A) along row i, yielding det(A). • If i=k, this represents the cofactor expansion along row k of a matrix whose k-th row has been replaced by row i. Such a matrix has two identical rows (row i and row k), so its determinant is zero! (This is known as expansion by alien cofactors). Thus (Aadj(A))ik=det(A)δik, where δik is the Kronecker delta. Therefore Aadj(A)=det(A)In. ■
Theorem (Cramer's Rule): Let Ax=b with det(A)=0. Then the unique solution coordinates are given by: xi=det(A)det(Ai(b)),i=1,…,n, where Ai(b) is the matrix obtained by replacing column i of A with the vector b.
⚠️ Common Traps & University Pitfalls · Determinants
Additivity Myth: In general, det(A+B)=det(A)+det(B). For instance, I2+(−I2)=0, yet det(I2)+det(−I2)=1+1=2=0.
Scalar Multiple Power Rule: For an n×n matrix, det(cA)=cndet(A), NOT cdet(A). For a 4×4 matrix, det(−A)=(−1)4det(A)=det(A).
Adjugate Transpose Transposition: The adjugate is adj(A)=CT. Forgetting to transpose the cofactor matrix produces incorrect inverse entries off the main diagonal.
Level 3 (Advanced) — Worked Example · Cramer's Rule & Invertibility Characterization
Problem: Solve the linear system for x2 using Cramer's Rule: 2x1+x2+x3x1−x2+2x33x1+2x2−x3=4=1=3
Let A be an invertible n×n matrix with n≥2. Prove analytically that: det(adj(A))=[det(A)]n−1.
Step 1 · Use the Fundamental Adjugate Identity:Aadj(A)=det(A)In.
Step 2 · Take determinants of both sides: Using the multiplicative property det(M1M2)=det(M1)det(M2): det(Aadj(A))=det(det(A)In).det(A)⋅det(adj(A))=[det(A)]ndet(In)=[det(A)]n(1)=[det(A)]n.
Step 3 · Divide by det(A): Since A is invertible, det(A)=0. Dividing both sides by det(A) yields: det(adj(A))=[det(A)]n−1.■
[Determinants & Geometric Volume · Practice 4] If the governing relation for this system is g(t)=6t+6, evaluate g(10) and find the inverse input t when g(t)=72.
Vector Spaces, Subspaces & The Four Fundamental Subspaces
📌 Linear Algebra · Chapter 5Structural Core · Vector space axioms & subspace criteria · Span, linear independence, basis & dimension · Rank-Nullity theorem & 4 fundamental subspaces
Abstract vector spaces extract the universal algebraic mechanisms of linear combinations beyond Euclidean space Rn to spaces of polynomials, matrices, and functions. This chapter develops the foundational hierarchy of linear structures: subspaces, span, linear independence, basis, dimension, and concludes with the Rank-Nullity Theorem and the Four Fundamental Subspaces of a matrix.
Level 1 · Foundations — Vector Space Axioms, Subspaces & The Subspace Test
A vector space is a set V equipped with addition + and scalar multiplication ⋅ over a field (here, R) satisfying ten fundamental closure and operational axioms.
Coordinate graph of f(x) = x² − 2 and its instantaneous tangent line at x = 1.
● Definition 5.1 — Abstract Vector Space & Subspace
A vector spaceV over R satisfies: 1. u+v∈V (Additive Closure). 2. u+v=v+u (Commutativity). 3. (u+v)+w=u+(v+w) (Associativity). 4. There exists 0∈V such that u+0=u (Additive Identity). 5. For each u∈V, there exists −u∈V such that u+(−u)=0 (Additive Inverse). 6. cu∈V for any scalar c∈R (Scalar Closure). 7. c(u+v)=cu+cv (Distributivity over Vectors). 8. (c+d)u=cu+du (Distributivity over Scalars). 9. c(du)=(cd)u (Scalar Compatibility). 10. 1u=u (Scalar Multiplicative Identity).
A subset W⊆V is a subspace if W is itself a vector space under the operations inherited from V.
● Theorem 5.1 & Proof · The 3-Step Subspace Test
Theorem: A non-empty subset W⊆V is a subspace of V if and only if: 1. 0∈W (Contains Zero Vector). 2. For all u,v∈W, u+v∈W (Closed under Addition). 3. For all u∈W and c∈R, cu∈W (Closed under Scalar Multiplication).
Proof: (⟹) If W is a subspace, it is a vector space in its own right, so axioms 1, 4, 6 guarantee conditions 1–3. (⟸) Assume conditions 1–3 hold. Closure under addition and scalar multiplication is given by conditions 2 and 3. Axioms 2, 3, 7, 8, 9, 10 hold automatically because elements of W are elements of V. Condition 1 gives the zero vector. Finally, for any u∈W, setting c=−1 in condition 3 gives −u=(−1)u∈W, providing additive inverses. Thus W satisfies all vector space axioms. ■
Level 1 (Foundations) — Worked Example · Applying the Subspace Test
Problem: Determine whether each subset is a subspace of R3: (a) W1={(x,y,z)∈R3:2x−3y+z=0} (b) W2={(x,y,z)∈R3:2x−3y+z=4} (c) W3={(x,y,z)∈R3:xy≥0}
Analysis: • (a) W1: (1) 2(0)−3(0)+0=0⟹0∈W1. (2) If u,v∈W1, then 2(u1+v1)−3(u2+v2)+(u3+v3)=(2u1−3u2+u3)+(2v1−3v2+v3)=0+0=0⟹u+v∈W1. (3) 2(cu1)−3(cu2)+(cu3)=c(2u1−3u2+u3)=c(0)=0⟹cu∈W1. W1 is a subspace (a plane through the origin). • (b) W2:2(0)−3(0)+0=0=4⟹0∈/W2. Fails immediately: W2 is NOT a subspace (an affine plane that misses the origin). • (c) W3: Contains 0. However, test closure under addition: u=(2,1,0)∈W3 (since (2)(1)=2≥0) and v=(−1,−3,0)∈W3 (since (−1)(−3)=3≥0). Their sum is u+v=(1,−2,0), but (1)(−2)=−2<0⟹u+v∈/W3. W3 is NOT a subspace.
Worked Example D — Euclidean Distance & Orthogonal Projection
A displacement has horizontal component Δx=5 and vertical component Δy=5. Find the squared magnitude d2 and the area of the right triangle formed with the axes.
Step-by-step Solution:
By the Pythagorean theorem, d2=(5)2+(5)2=25+25=50.
The enclosed right-triangle area is A=21(5)(5)=12.5.
Level 1 · Foundations
Let M2×2 be the vector space of all 2×2 real matrices. Show that the set of all symmetric matrices W={A∈M2×2:AT=A} is a subspace of M2×2.
Step 1 · Zero vector: The 2×2 zero matrix satisfies 0T=0, so 0∈W.
Step 2 · Closed under addition: Let A,B∈W. Then AT=A and BT=B. Using transpose properties: (A+B)T=AT+BT=A+B⟹A+B∈W.
Step 3 · Closed under scalar multiplication: Let A∈W and c∈R: (cA)T=cAT=cA⟹cA∈W. All three conditions hold, so W is a subspace of M2×2.
Level 2 · Medium — Span, Linear Independence, Basis & Dimension
The structural skeleton of any vector space is defined by linear independence and bases. A basis is a minimal spanning set and a maximal linearly independent set, providing unique coordinates for every vector in the space.
1. Span: The set of all linear combinations of vectors {v1,…,vk}: span(v1,…,vk)={i=1∑kcivi:ci∈R}. The span is always a subspace of V.
2. Linear Independence: Vectors {v1,…,vk} are linearly independent if: c1v1+c2v2+⋯+ckvk=0⟹c1=c2=⋯=ck=0. If a non-trivial linear combination yields 0, the set is linearly dependent.
3. Basis & Dimension: A subset B={b1,…,bn}⊂V is a basis for V if: • B is linearly independent. • span(B)=V. The number of vectors n in any basis of V is an invariant, called the dimensiondim(V).
Theorem: Let B={b1,…,bn} be a basis for vector space V. Then every vector v∈V can be written in exactly one unique way as a linear combination of the basis vectors: v=c1b1+c2b2+⋯+cnbn.
Proof: Since B spans V, there exists at least one representation v=∑i=1ncibi. Suppose there exists a second representation v=∑i=1ndibi. Subtracting the two equations gives: 0=v−v=i=1∑n(ci−di)bi. Because B is linearly independent, the only solution to this linear combination is the trivial one: ci−di=0⟹ci=difor all i=1,…,n. Hence the coefficients are uniquely determined. The tuple [v]B=(c1,…,cn)T constitutes the coordinate vector of v relative to B. ■
⊙ Explorer · Span of two vectors in ℝ²interactive
u, v, the coefficients, and the resulting linear combination
quantity
value
Drag to rotate, scroll to zoom · loads a small 3D library on first use
c₁u + c₂v⟨3.00, 1.73⟩
span(u, v)all of ℝ²
u = ⟨2, 0⟩ is fixed; v's direction is set by the angle slider. When the angle is between 1° and 179°, u and v are linearly independent and span all of ℝ². When the angle is 0° or 180°, u and v are collinear, collapsing the span to a single line.
Comparison of a linear subspace and an affine line in R2. The line through the origin y=0.5x contains the zero vector and is closed under addition and scalar multiplication; the shifted line y=0.5x+2 fails the zero vector axiom.
Level 2 (Medium) — Worked Example · Finding Bases for Null Space & Column Space
Problem: Find a basis for the column space Col(A) and null space Null(A) of: A=121−2−4−2011396.
Step 1 · Row Reduce A to RREF:121−2−4−2011396R2←R2−2R1R3←R3−R1100−200011333R3←R3−R2100−200010330.
Step 2 · Basis for Column Space Col(A): Pivots occur in columns 1 and 3. The basis for Col(A) consists of the corresponding original columns of A (never the RREF columns!): Basis(Col(A))=⎩⎨⎧121,011⎭⎬⎫,dim(Col(A))=2.
Step 3 · Basis for Null Space Null(A): Columns 2 and 4 lack pivots ⟹x2=s,x4=t free. Basic variables: x1=2s−3t, x3=−3t. x=2s−3ts−3tt=s2100+t−30−31.Basis(Null(A))=⎩⎨⎧2100,−30−31⎭⎬⎫,dim(Null(A))=2.
Level 2 · Medium
Determine whether the polynomials p1(x)=1+x, p2(x)=x+x2, and p3(x)=1+x2 form a basis for P2 (the vector space of polynomials of degree at most 2).
Step 1 · Coordinate vectors: The standard basis for P2 is E={1,x,x2}, so dim(P2)=3. The coordinate vectors are: [p1]E=110,[p2]E=011,[p3]E=101.
Step 2 · Test independence via determinant:det110011101=1(1−0)−0+1(1−0)=1+1=2=0. Since the determinant is non-zero, the three coordinate vectors are linearly independent. Because dim(P2)=3, any three linearly independent vectors form a basis. Yes, {p1,p2,p3} is a basis for P2.
Level 3 · Advanced / University Rigor — The Four Fundamental Subspaces & The Rank-Nullity Theorem
Every m×n matrix A determines four fundamental subspaces that completely characterize the geometry of linear mappings and matrix equations.
● Theorem 5.3 & Proof · The Rank-Nullity Theorem
Theorem (Rank-Nullity / Dimension Theorem): Let A be an m×n matrix. Then: rank(A)+nullity(A)=n, where rank(A)=dim(Col(A)) and nullity(A)=dim(Null(A)).
Proof via RREF: Let R be the reduced row echelon form of A. Row operations preserve linear relationships among columns, so the pivot columns of A form a basis for Col(A). Therefore, rank(A)=r, where r is the number of pivot columns. The remaining n−r non-pivot columns correspond directly to the free variables in the homogeneous equation Ax=0. Each free variable xj produces exactly one special basis vector in the parametric representation of Null(A), and these n−r vectors are linearly independent because each vector has a 1 in its respective free variable coordinate and 0 in all other free coordinates. Thus nullity(A)=n−r. Adding the two dimensions: rank(A)+nullity(A)=r+(n−r)=n.■
● Theorem 5.4 · The Fundamental Theorem of Linear Algebra (Gilbert Strang)
For any m×n matrix A with rank r:
Subspace
Notation
Ambient Space
Dimension
Orthogonal Complement
Column Space
C(A)=Col(A)
Rm
r
N(AT)
Null Space
N(A)=Null(A)
Rn
n−r
C(AT)
Row Space
C(AT)=Row(A)
Rn
r
N(A)
Left Null Space
N(AT)
Rm
m−r
C(A)
Orthogonal Complements:Null(A)=[Row(A)]⊥,Null(AT)=[Col(A)]⊥. This implies the orthogonal direct sum decompositions: Rn=Row(A)⊕Null(A),Rm=Col(A)⊕Null(AT).
⚠️ Common Traps & University Pitfalls · Vector Spaces
Basis for Column Space: You must take the pivot columns from the original matrix A, NOT from its row reduced form R. Row operations change the column space! By contrast, a basis for the row space can be taken directly from the non-zero rows of R.
Rank vs Dimension of Ambient Space: Rank is the dimension of the column space (number of pivots), bounded by min(m,n). A 3×5 matrix can have rank at most 3, so its nullity must be at least 5−3=2.
Union of Subspaces: The union W1∪W2 of two subspaces is almost never a subspace (it is a subspace if and only if one is contained in the other). The sum W1+W2={w1+w2:w1∈W1,w2∈W2} is always a subspace.
Level 3 (Advanced) — Worked Example · Complete Four Fundamental Subspaces Decomposition
Problem: For A=12−136−3393274, determine the dimension and find a basis for each of the four fundamental subspaces.
Step 1 · Row Reduce A to RREF:12−136−3393274R2←R2−2R1R3←R3+R1100300336236R3←R3−2R2100300310210R1←R1−3R2100300010−110. Pivots in columns 1 and 3 ⟹rank(A)=r=2.
Step 2 · The Four Subspaces: • Col(A)⊂R3: Original columns 1 and 3: Basis=⎩⎨⎧12−1,393⎭⎬⎫,dim=2. • Null(A)⊂R4: Free variables x2=s,x4=t. x1=−3s+t, x3=−t: Basis=⎩⎨⎧−3100,10−11⎭⎬⎫,dim=4−2=2. • Row(A)⊂R4: Non-zero rows of RREF: Basis=⎩⎨⎧130−1,0011⎭⎬⎫,dim=2. • Null(AT)⊂R3: Left null space has dimension m−r=3−2=1. Solve ATy=0: 13322697−1−334→10000100−5200⟹y1=5t,y2=−2t,y3=t.Basis=⎩⎨⎧5−21⎭⎬⎫,dim=1.
Let A be a 5×8 matrix with rank(A)=4. (a) What is the dimension of the solution space to Ax=0? (b) Is Ax=b consistent for every b∈R5? Explain. (c) What is the dimension of Null(AT)?
(a) Null Space Dimension: By the Rank-Nullity Theorem, rank(A)+nullity(A)=n. Here n=8 and rank(A)=4: nullity(A)=8−4=4.
(b) Consistency for all b:No.Col(A) has dimension 4, which is a proper subspace of R5 (dimension 5). Therefore, there exist vectors b∈R5 that do not lie in Col(A), making Ax=b inconsistent.
(c) Left Null Space Dimension: The left null space Null(AT) is the orthogonal complement of Col(A) in Rm=R5. Its dimension is: dim(Null(AT))=m−rank(A)=5−4=1.
[Vector Spaces, Subspaces & The Four Fundamental Subspaces · Practice 4] If the governing relation for this system is g(t)=6t+6, evaluate g(10) and find the inverse input t when g(t)=72.
📌 Linear Algebra · Chapter 6Spectral Theory · Characteristic equation, trace & determinant relations · Multiplicity, eigenvector independence & A = PDP⁻¹ · Matrix powers, Cayley–Hamilton & systems of ODEs
Under a general linear transformation, vectors typically experience both stretching and rotation. An eigenvector identifies an invariant direction that is purely scaled by a scalar factor known as the eigenvalue. Spectral theory decouples coupled linear systems, enabling explicit computation of matrix powers, Markov processes, and systems of linear differential equations.
Let A∈Rn×n. A non-zero vector v∈Rn is an eigenvector of A corresponding to eigenvalue λ∈C if: Av=λv(v=0).
Coordinate graph of f(x) = x² − 2 and its instantaneous tangent line at x = 1.
● Definition 6.1 — Eigenspaces & The Characteristic Polynomial
1. Eigenspace: For a given eigenvalue λ, the set of all eigenvectors along with the zero vector forms a subspace of Rn, the eigenspaceEλ: Eλ=Null(A−λI)={v∈Rn:(A−λI)v=0}.
2. The Characteristic Equation: The homogeneous system (A−λI)v=0 has a non-trivial solution v=0 if and only if A−λI is singular (non-invertible): p(λ)=det(A−λI)=0. Expanding the determinant yields a degree-n polynomial in λ, the characteristic polynomial.
3. Trace & Determinant Invariants: If λ1,…,λn are the eigenvalues of A counted with multiplicity: i=1∑nλi=tr(A)=i=1∑naii,i=1∏nλi=det(A).
Level 1 (Foundations) — Worked Example · Computing Eigenvalues, Eigenspaces & Invariant Checks
Problem: Find the eigenvalues and corresponding eigenspace bases for A=(4213), and verify the trace and determinant relations.
Worked Example D — Euclidean Distance & Orthogonal Projection
A displacement has horizontal component Δx=5 and vertical component Δy=5. Find the squared magnitude d2 and the area of the right triangle formed with the axes.
Step-by-step Solution:
By the Pythagorean theorem, d2=(5)2+(5)2=25+25=50.
The enclosed right-triangle area is A=21(5)(5)=12.5.
Level 1 · Foundations
Find the eigenvalues of the upper-triangular matrix: T=600−2−30514.
The characteristic matrix is: T−λI=6−λ00−2−3−λ0514−λ. Since T−λI is upper-triangular, its determinant is the product of its diagonal entries: det(T−λI)=(6−λ)(−3−λ)(4−λ)=0. Thus the eigenvalues are the diagonal entries themselves: λ1=6,λ2=−3,λ3=4.
Level 2 · Medium — Linear Independence of Eigenvectors & Matrix Diagonalization
When a matrix has a full set of linearly independent eigenvectors, it can be factored into a product A=PDP−1 where D is diagonal. This diagonal representation decouples the action of A into independent 1D scalings along the eigenvector coordinate axes.
● Theorem 6.1 & Analytical Proof · Independence of Eigenvectors for Distinct Eigenvalues
Theorem: Let v1,v2,…,vk be eigenvectors of A corresponding to mutually distinct eigenvalues λ1,λ2,…,λk (i.e. λi=λj for i=j). Then the set {v1,…,vk} is linearly independent.
Proof by Induction: • Base Case (k=1):v1=0 by definition of an eigenvector, so {v1} is linearly independent. • Inductive Step: Assume the statement holds for k−1 eigenvectors. Suppose: c1v1+c2v2+⋯+ckvk=0.(Equation 1) Multiply both sides by λk: c1λkv1+c2λkv2+⋯+ckλkvk=0.(Equation 2) Apply the matrix A to Equation 1, using Avi=λivi: c1λ1v1+c2λ2v2+⋯+ckλkvk=0.(Equation 3) Subtract Equation 2 from Equation 3: c1(λ1−λk)v1+c2(λ2−λk)v2+⋯+ck−1(λk−1−λk)vk−1+0=0. By the inductive hypothesis, {v1,…,vk−1} is linearly independent, so each coefficient must be zero: ci(λi−λk)=0for i=1,…,k−1. Because all eigenvalues are distinct, λi−λk=0, forcing ci=0 for all i=1,…,k−1. Substituting these zeros back into Equation 1 gives ckvk=0, which forces ck=0 since vk=0. Therefore, all ci=0, proving linear independence. ■
● Theorem 6.2 · The Diagonalization Theorem
An n×n matrix A is diagonalizable (there exists invertible P and diagonal D such that A=PDP−1) if and only if A has n linearly independent eigenvectors.
In this case: P=[v1v2…vn],D=λ10⋮00λ2⋮0……⋱…00⋮λn, where column j of P is an eigenvector corresponding to the diagonal entry λj in D.
Corollaries: • If A has n distinct eigenvalues, A is automatically diagonalizable. • Powers of A: For any integer k≥1, Ak=PDkP−1=Pλ1k0⋱0λnkP−1.
⊙ Explorer · Eigenvectors — invariant transformation directionsinteractive
The matrix, its eigenvalues, and an eigenvector for each
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λ₁5.000
λ₂2.000
The grey curve is the unit circle; indigo is its image ellipse under A. The amber and teal lines represent the real eigenvector directions along which the transformation purely dilates vectors without rotation. When the discriminant tr(A)² - 4 det(A) < 0, eigenvalues become complex and no real invariant line exists.
Eigenspaces of A=(2112). Vectors lying along y=x are scaled by λ1=3; vectors along y=−x are scaled by λ2=1. The directions of the eigenvectors remain invariant under transformation by A.
Level 2 (Medium) — Worked Example · High Matrix Power via Diagonalization
Problem: Compute A10 for A=(1331) using diagonalization.
Explain why the matrix A=(3013) is not diagonalizable.
Step 1 · Algebraic Multiplicity:A is upper-triangular, so its characteristic polynomial is p(λ)=(3−λ)2=0. The only eigenvalue is λ=3 with algebraic multiplicity ma(3)=2.
Step 2 · Geometric Multiplicity: Compute the eigenspace E3=Null(A−3I): A−3I=(0010). This matrix has pivot in column 2, with only one free variable x1=t. The eigenvectors are v=t(10), so dim(E3)=1.
Conclusion: Since the geometric multiplicity mg(3)=1 is strictly less than the algebraic multiplicity ma(3)=2, A fails to possess 2 linearly independent eigenvectors. Therefore, A is defective and NOT diagonalizable.
Level 3 · Advanced / University Rigor — The Cayley–Hamilton Theorem & Matrix Differential Equations
The Cayley–Hamilton Theorem asserts that every square matrix satisfies its own characteristic equation, providing a powerful analytic tool for computing inverse matrices and powers. Furthermore, spectral theory solves systems of first-order linear differential equations x′(t)=Ax(t) via the matrix exponential.
● Theorem 6.3 & Proof · The Cayley–Hamilton Theorem
Theorem (Cayley–Hamilton): Let A be an n×n matrix with characteristic polynomial p(λ)=det(A−λI)=(−1)nλn+cn−1λn−1+⋯+c1λ+c0. Then substituting A into its own polynomial yields the zero matrix: p(A)=(−1)nAn+cn−1An−1+⋯+c1A+c0I=0.
Proof via Adjugate Matrix: Recall from Theorem 4.2 that for any square matrix, Madj(M)=det(M)I. Setting M=A−λI: (A−λI)adj(A−λI)=det(A−λI)I=p(λ)I. The entries of adj(A−λI) are cofactors of A−λI, which are polynomials in λ of degree at most n−1. Therefore, adj(A−λI) can be expanded as a matrix polynomial: adj(A−λI)=Bn−1λn−1+Bn−2λn−2+⋯+B1λ+B0, where each Bk∈Rn×n. Substituting into the adjugate identity: (A−λI)(Bn−1λn−1+⋯+B0)=(cnλn+cn−1λn−1+⋯+c0)I. Expanding and equating coefficients of λk yields a telescoping sequence of matrix equations. Pre-multiplying the k-th equation by Ak and summing all equations eliminates all terms on the left side, yielding 0=p(A). ■
⚠️ Common Traps & University Pitfalls · Eigenvalues & Diagonalization
Row Operations Alter Eigenvalues: Performing elementary row operations A→R preserves row space and null space, but completely destroys eigenvalues: det(A−λI)=det(R−λI). Never row reduce A prior to setting up det(A−λI)=0!
Defective Matrices: Having repeated eigenvalues (ma(λ)≥2) does NOT automatically mean the matrix is not diagonalizable. For example, I2 has λ=1 with multiplicity 2, but E1=R2 has dimension 2, so I2 is diagonalizable. A matrix is non-diagonalizable only when mg(λ)<ma(λ) for some eigenvalue.
Complex Eigenvalues for Real Matrices: A real matrix can have complex conjugate eigenvalues a±bi representing rotation-dilations. A real matrix with complex eigenvalues is not diagonalizable over R (though it is diagonalizable over C).
Level 3 (Advanced) — Worked Example · Systems of Differential Equations via Diagonalization
Problem: Solve the initial value problem for the coupled linear ODE system: x′(t)=(1221)x(t),x(0)=(51).
Step 1 · Diagonalize Coefficient Matrix A:det(A−λI)=(1−λ)2−4=λ2−2λ−3=(λ−3)(λ+1)=0⟹λ1=3,λ2=−1. • For λ1=3: A−3I=(−222−2)⟹v1=(11). • For λ2=−1: A+I=(2222)⟹v2=(1−1).
Step 2 · Decouple via Coordinate Change x=Py: Since A=PDP−1, x′=Ax⟹Py′=PDy⟹y′=Dy. The decoupled solutions are y1(t)=c1e3t and y2(t)=c2e−t. Therefore: x(t)=c1e3tv1+c2e−tv2=c1e3t(11)+c2e−t(1−1).
Step 2 · Apply Cayley–Hamilton: A2−5A+6I=0: From this: A2=5A−6I. Multiply by A: A3=5A2−6A=5(5A−6I)−6A=25A−30I−6A=19A−30I. Multiply by A again: A4=19A2−30A=19(5A−6I)−30A=95A−114I−30A=65A−114I.
Step 3 · Express A−1: Multiply A2−5A+6I=0 by A−1: A−5I+6A−1=0⟹6A−1=5I−A⟹A−1=−61A+65I.
[Eigenvalues, Eigenvectors & Diagonalization · Practice 4] If the governing relation for this system is g(t)=6t+6, evaluate g(10) and find the inverse input t when g(t)=72.
Linear Transformations, Matrix Representations & Change of Basis
A function between vector spaces that respects addition and scalar multiplication is called a linear transformation. While Chapter 3 introduced matrix–vector multiplication as an algebraic operation, this chapter reveals the deeper geometric reality: every linear operator on finite-dimensional spaces is intrinsically represented by a matrix once bases are chosen. Linear transformations form the conceptual bridge connecting abstract vector spaces, geometry, and linear operator theory across quantum mechanics, computer graphics, and continuous dynamical systems.
● Level 1 Foundation — Definition 7.1: Linear Transformation
Let V and W be vector spaces over R. A mapping T:V→W is a linear transformation (or linear operator when V=W) if and only if for all vectors u,v∈V and every scalar c∈R:
Additivity:T(u+v)=T(u)+T(v)
Scalar Homogeneity:T(cu)=cT(u)
These two conditions are equivalent to the single superposition condition: T(c1u+c2v)=c1T(u)+c2T(v) for all scalars c1,c2. By induction, a linear map preserves arbitrary linear combinations: T(i=1∑kcivi)=i=1∑kciT(vi).
Zero Vector Property: Taking c=0 yields T(0V)=T(0⋅0V)=0⋅T(0V)=0W. Every linear map strictly maps the origin of the domain to the origin of the codomain. Any mapping with T(0)=0 (such as the affine shift f(x)=Ax+b with b=0) is not linear.
Coordinate graph of f(x) = x² − 2 and its instantaneous tangent line at x = 1.
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Numerical table for Linear Transformations, Matrix Representations & Change of Basis
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Primary Metric2.00
Secondary Metric1.00
▲ Level 1 Foundation — Theorem 7.1: The Standard Matrix Representation
Let T:Rn→Rm be a linear transformation. Then there exists a unique m×n matrix [T] (or A), called the standard matrix of T, such that: T(x)=[T]xfor all x∈Rn. Furthermore, the columns of [T] are precisely the images of the standard basis vectors e1,e2,…,en of Rn: [T]=∣T(e1)∣∣T(e2)∣⋯∣T(en)∣.
Proof of Theorem 7.1
Any vector x=(x1,x2,…,xn)T∈Rn can be uniquely expressed as a linear combination of standard basis vectors: x=x1e1+x2e2+⋯+xnen=j=1∑nxjej. Applying the linear transformation T and utilizing linearity across the summation and scalar coefficients: T(x)=T(j=1∑nxjej)=j=1∑nxjT(ej). By the fundamental definition of matrix–vector multiplication (a linear combination of the matrix columns weighted by the vector components): j=1∑nxjT(ej)=∣T(e1)∣∣T(e2)∣⋯∣T(en)∣x1x2⋮xn=[T]x. Uniqueness follows from evaluating T(ej): if Ax=Bx for all x, setting x=ej yields Aej=Bej, forcing the j-th columns of A and B to be identical for all j. Hence [T] is unique. ■
Worked Example 7.A (Level 1) — Canonical Geometric Operators in ℝ²
Construct the standard transformation matrices for three foundational geometric operators on R2:
Counterclockwise Rotation by Angle θ (Rθ): Under rotation by θ, e1=(1,0)T↦(cosθ,sinθ)T and e2=(0,1)T↦(−sinθ,cosθ)T. Therefore: [Rθ]=(cosθsinθ−sinθcosθ).
Reflection Across a Line Through the Origin at Angle α (Mα): Unit vector along the line is u=(cosα,sinα)T. Reflection maps x↦2(x⋅u)u−x. Evaluating on standard basis vectors: [Mα]=(cos(2α)sin(2α)sin(2α)−cos(2α)).
Horizontal Shear by Factor k (Sk): Sk(x,y)=(x+ky,y). Here Sk(e1)=(1,0)T and Sk(e2)=(k,1)T, giving: [Sk]=(10k1).
Kernel (Null Space): The set of all input vectors mapped to the zero vector in W: ker(T)=null(T)={v∈V:T(v)=0W}.ker(T) is a vector subspace of the domain V. Its dimension is the nullity of T: nullity(T)=dim(kerT).
Image (Range): The set of all output vectors achievable in W: im(T)=range(T)={w∈W:∃v∈V such that T(v)=w}.im(T) is a vector subspace of the codomain W, equivalent to the column space Col([T]). Its dimension is the rank of T: rank(T)=dim(imT).
Injectivity (One-to-One):T is injective if T(u)=T(v)⟹u=v.
Surjectivity (Onto):T is surjective if im(T)=W (i.e., rank(T)=dim(W)).
Isomorphism (Bijective):T is an isomorphism if it is both injective and surjective. In this case, V and W are isomorphic (V≅W), which requires dim(V)=dim(W).
▲ Level 2 Analytical Application — Theorem 7.2: Injectivity and the Trivial Kernel
Let T:V→W be a linear transformation. Then T is injective (one-to-one) if and only if ker(T)={0V} (i.e., nullity(T)=0).
Proof of Theorem 7.2
Forward Direction (⟹): Assume T is injective. We already know T(0V)=0W, so 0V∈ker(T). Now let v∈ker(T). Then T(v)=0W. Thus T(v)=T(0V). Because T is injective, this directly implies v=0V. Hence, ker(T)={0V}.
Reverse Direction (⟸): Assume ker(T)={0V}. Suppose T(u)=T(v) for some u,v∈V. Subtracting T(v) from both sides and applying linearity gives: T(u)−T(v)=T(u−v)=0W. By definition of the kernel, u−v∈ker(T). Since ker(T)={0V}, we must have u−v=0V, which means u=v. Therefore, T is one-to-one. ■
▲ Level 2 Analytical Application — Theorem 7.3: The Dimension Theorem for Linear Maps
If V is a finite-dimensional vector space and T:V→W is a linear transformation, then: dim(kerT)+dim(imT)=dim(V), or equivalently: nullity(T)+rank(T)=dim(V).
Proof of the Dimension Theorem (Rank-Nullity)
Let n=dim(V) and let k=dim(kerT). If k=0, the kernel is {0} with basis ∅. Otherwise, choose a basis BK={u1,u2,…,uk} for ker(T).
By the Basis Extension Theorem, extend BK to a complete basis for V: B={u1,…,uk,w1,…,wn−k}. We claim that the set S={T(w1),T(w2),…,T(wn−k)} is a basis for im(T).
1. S spans im(T): Let y∈im(T). Then y=T(v) for some v∈V. Express v in terms of B: v=i=1∑kciui+j=1∑n−kdjwj. Applying T and noting T(ui)=0 for all i: y=T(v)=i=1∑kciT(ui)+j=1∑n−kdjT(wj)=j=1∑n−kdjT(wj). Thus span(S)=im(T).
2. S is linearly independent: Suppose j=1∑n−kdjT(wj)=0W. By linearity, T(∑j=1n−kdjwj)=0W, which proves that the vector w=∑j=1n−kdjwj lies in ker(T). Since BK is a basis for ker(T), w can be written as: j=1∑n−kdjwj=i=1∑kciui⟺i=1∑k(−ci)ui+j=1∑n−kdjwj=0V. Because the combined set B is linearly independent in V, all coefficients must vanish: c1=⋯=ck=d1=⋯=dn−k=0. In particular, d1=d2=⋯=dn−k=0, establishing that S is linearly independent.
Since S spans im(T) and is linearly independent, it is a basis for im(T). Hence dim(imT)=n−k=dim(V)−dim(kerT), completing the proof. ■
Worked Example 7.B (Level 2) — Dimension Theorem & Kernel Computation
Consider the linear transformation T:R4→R3 defined by T(x)=Ax where: A=124−120204328. Determine a basis for ker(T), a basis for im(T), and verify the Dimension Theorem.
Step 1: Row reduce A to RREF:124−120204328R2−2R1,R3−4R1100−1442−4−43−4−4R3−R2,R2/4100−1102−103−10R1+R21000101−102−10.
Step 2: Basis for ker(T): Pivot variables are x1,x2; free variables are x3=s,x4=t. Expressing pivot variables: x1=−s−2t and x2=s+t. In parametric vector form: x=s−1110+t−2101. Thus a basis for ker(T) is ⎩⎨⎧−1110,−2101⎭⎬⎫, and nullity(T)=2.
Step 3: Basis for im(T): Pivot columns are columns 1 and 2 of the original matrix A: Basis for im(T)=⎩⎨⎧124,−120⎭⎬⎫,rank(T)=2.
● Level 3 University Rigor — Definition 7.3: Matrix Relative to Arbitrary Bases
Let V and W be vector spaces with ordered bases B={v1,v2,…,vn} and C={w1,w2,…,wm}, respectively. Let [v]B denote the coordinate vector of v relative to B.
The matrix representation of T:V→W relative to bases B and C, denoted [T]BC (or simply [T]B when V=W and B=C), is the m×n matrix whose j-th column is the coordinate vector of T(vj) with respect to C: [T]BC=∣[T(v1)]C∣∣[T(v2)]C∣⋯∣[T(vn)]C∣. This matrix satisfies the commutative diagram property: [T(x)]C=[T]BC[x]Bfor all x∈V.
▲ Level 3 University Rigor — Theorem 7.4: Change of Basis & Similarity Transformations
Let T:V→V be a linear operator on an n-dimensional space V, and let B and C be two ordered bases of V. Let P=PB←C be the change-of-basis transition matrix from C to B, whose columns are [wj]B for wj∈C.
Then P is invertible, [x]B=P[x]C, and the matrices representing T under B and C are related by a similarity transformation: [T]C=P−1[T]BP.
Similarity Invariants: Because [T]C and [T]B are similar matrices (B=P−1AP), they share identical intrinsic geometric properties independent of the chosen coordinate basis:
det([T]C)=det([T]B)
tr([T]C)=tr([T]B)
p[T]C(λ)=p[T]B(λ) (identical characteristic polynomial and eigenvalues)
Identical rank and nullity
Proof of Theorem 7.4
Let x∈V. By definition of the transition matrix, coordinates convert via: [x]B=P[x]C⟺[x]C=P−1[x]B. Applying this conversion rule to the transformed vector T(x): [T(x)]C=P−1[T(x)]B. Substituting the definition [T(x)]B=[T]B[x]B: [T(x)]C=P−1([T]B[x]B)=P−1[T]B(P[x]C)=(P−1[T]BP)[x]C. Since this equality holds for all coordinate vectors [x]C∈Rn, uniqueness of the matrix representation establishes: [T]C=P−1[T]BP.■
Worked Example 7.C (Level 3) — Differential Operator Representation in Polynomial Spaces
Let P2 be the vector space of real polynomials of degree at most 2. Let D:P2→P2 be the linear differential operator D(p(x))=xp′(x)−2p(x).
Find the matrix representation [D]B relative to the standard monomial basis B={1,x,x2}.
Find ker(D) and im(D), and determine whether D is an isomorphism.
Step 1: Compute images of basis elements:
For p1(x)=1: D(1)=x(0)−2(1)=−2=−2(1)+0(x)+0(x2)⟹[D(1)]B=−200.
For p2(x)=x: D(x)=x(1)−2(x)=−x=0(1)−1(x)+0(x2)⟹[D(x)]B=0−10.
For p3(x)=x2: D(x2)=x(2x)−2(x2)=0=0(1)+0(x)+0(x2)⟹[D(x2)]B=000.
Assembling the columns into the matrix representation: [D]B=−2000−10000.
Step 2: Kernel and Image Analysis: The matrix has a pivot in columns 1 and 2, while column 3 is a free column. The null space of [D]B is spanned by (0,0,1)T, corresponding to the polynomial x2. Therefore: ker(D)=span{x2},dim(kerD)=1. Since ker(D)={0}, D is not injective, and by Rank-Nullity rank(D)=3−1=2<3 (not surjective). Hence D is not an isomorphism. The image is spanned by {p1,p2}: im(D)=span{1,x}.
⚠️ Common Traps & University Pitfalls — Linear Transformations
Affine Shifts are NOT Linear: A common exam trap is mistaking T(x)=Ax+b (for b=0) for a linear transformation. Because T(0)=b=0, additivity and scalar scaling fail completely. It is an affine map, never a linear transformation.
Codomain Dimension vs. Range Dimension: If T:R5→R3, students frequently confuse the codomain R3 with im(T). The dimension of the range is rank(T)≤3; T is onto only if rank(T)=3.
Inverting the Transition Matrix Direction: In changing basis with [T]C=P−1[T]BP, the transition matrix P=PB←C has columns given by the coordinates of the new basis C relative to the old basis B. Swapping P and P−1 flips the entire coordinate perspective.
Composition Order: The composite transformation (S∘T)(x)=S(T(x)) corresponds to matrix product [S][T], applied from right to left. Reversing this product to [T][S] computes T∘S, which is generally completely distinct or undefined when dimensions differ.
Linearity Axioms
T(c1u+c2v)=c1T(u)+c2T(v)
Standard Matrix
Columns are T(e1),…,T(en)
Dimension Theorem
dim(kerT)+dim(imT)=dim(V)
Change of Basis
[T]C=P−1[T]BP
Geometric effect of the rotation transformation T(x)=(01−10)x. The standard basis vector e1=⟨1,0⟩ maps to ⟨0,1⟩, rotating the plane 90∘ counterclockwise.
Practice Problem (Level 1) — Composition & Matrix Inverse
T:R2→R2 projects every vector onto the x-axis: T(x,y)=(x,0). (a) Verify T is linear and find its standard matrix [T]. (b) Find ker(T) and im(T), and verify the Dimension Theorem. (c) Explain geometrically and algebraically why [T] cannot be invertible.
Show solution
(a) T(c1(x1,y1)+c2(x2,y2))=T(c1x1+c2x2,c1y1+c2y2)=(c1x1+c2x2,0)=c1(x1,0)+c2(x2,0)=c1T(x1,y1)+c2T(x2,y2) — strictly linear. Testing standard basis: T(1,0)=(1,0) and T(0,1)=(0,0), hence [T]=(1000).
(b) T(x,y)=(0,0)⟺x=0. Thus ker(T)={(0,y):y∈R}=span((0,1)T), which is the y-axis (dim=1). The range is im(T)={(x,0):x∈R}=span((1,0)T), which is the x-axis (dim=1). Check: dim(kerT)+dim(imT)=1+1=2=dim(R2). ✓
(c) Algebraically, det([T])=(1)(0)−(0)(0)=0, so [T] is singular. Geometrically, projection permanently collapses the entire 2D plane onto a 1D line; all information along the y-axis is lost (kerT={0}), so the operation is non-injective and cannot be uniquely reversed.
Worked Example · Matrix Representation Under Non-Standard Basis
Let T:R2→R2 be reflection across the line y=2x. Find [T]B relative to the basis B={v1,v2} where v1=(12) lies on the line and v2=(−21) is orthogonal to the line.
Solution: Since v1 lies along the reflection line, T(v1)=v1=1v1+0v2⟹[T(v1)]B=(10). Since v2 is perpendicular to the line, reflection negates it: T(v2)=−v2=0v1−1v2⟹[T(v2)]B=(0−1). Therefore: [T]B=(100−1). (Notice how selecting an eigenbasis completely diagonalizes the geometric transformation!)
Worked Example · Injectivity & Codomain Dimensions
Can a linear transformation T:R5→R3 be injective (one-to-one)? Can it be surjective (onto)?
Solution: By the Dimension Theorem, dim(kerT)+dim(imT)=5. Since im(T)⊆R3, dim(imT)≤3. Thus: dim(kerT)=5−dim(imT)≥5−3=2. Since dim(kerT)≥2>0, ker(T)={0}, so Tcannot be one-to-one under any circumstances. However, if rank(T)=3=dim(R3), Tcan be surjective.
Worked Example D — Euclidean Distance & Orthogonal Projection
A displacement has horizontal component Δx=5 and vertical component Δy=5. Find the squared magnitude d2 and the area of the right triangle formed with the axes.
Step-by-step Solution:
By the Pythagorean theorem, d2=(5)2+(5)2=25+25=50.
The enclosed right-triangle area is A=21(5)(5)=12.5.
A linear transformation T:R7→R4 has dim(kerT)=3. What is the dimension of the range of T, and is T surjective?
By the Dimension Theorem, nullity(T)+rank(T)=dim(domain). Here: 3+rank(T)=7⟹rank(T)=dim(imT)=4. Since dim(imT)=4=dim(R4), the range spans the entire codomain, so T is surjective (onto).
Find the standard matrix [T] of the linear operator T:R2→R2 that first rotates vectors by 90∘ counterclockwise, and then reflects the result across the line y=x.
Step 1: Track images of standard basis vectors: For e1=(10): Rotation by 90∘ CCW sends e1 to (01). Reflection across y=x swaps coordinates: (01)⟶(10). For e2=(01): Rotation by 90∘ CCW sends e2 to (−10). Reflection across y=x swaps coordinates: (−10)⟶(0−1).
Step 2: Assemble standard matrix: The columns of [T] are T(e1) and T(e2): [T]=(100−1). Alternatively, via matrix multiplication of standard operators: [T]=[Mπ/4][Rπ/2]=(0110)(01−10)=(100−1). This composite operator is geometrically equivalent to reflection across the x-axis!
[Linear Transformations, Matrix Representations & Change of Basis · Practice 3] If the governing relation for this system is g(t)=5t+5, evaluate g(8) and find the inverse input t when g(t)=50.
[Linear Transformations, Matrix Representations & Change of Basis · Practice 4] If the governing relation for this system is g(t)=6t+6, evaluate g(10) and find the inverse input t when g(t)=72.
Orthogonality, Gram–Schmidt, QR Decomposition & Least Squares
Orthogonality is the central organizing principle of modern computational linear algebra. By decomposing vector spaces into mutually perpendicular subspaces, orthogonality transforms complex matrix inversions into simple inner products, eliminates numerical instability, and powers the Gram–Schmidt process and QR decomposition. Furthermore, orthogonal projections resolve overdetermined, inconsistent linear systems via the method of least squares — the mathematical engine underpinning linear regression, statistical estimation, signal processing, and machine learning.
● Level 1 Foundation — Definition 8.1: Orthogonal & Orthonormal Systems
Let V be an inner product space (such as Rn with standard Euclidean inner product ⟨u,v⟩=u⋅v).
Two vectors u,v are orthogonal (denoted u⊥v) if and only if u⋅v=0.
A set of vectors S={u1,u2,…,uk} is an orthogonal set if ui⋅uj=0 whenever i=j.
An orthogonal set S is orthonormal if every vector has unit length: ui⋅uj=δij={10if i=jif i=j where δij is the Kronecker delta.
An orthonormal basis (ONB) for a subspace W is a basis of W that is orthonormal.
Orthogonal Fourier Expansion: If {q1,…,qn} is an orthonormal basis for Rn, any vector x expands effortlessly without solving a system of equations: x=(x⋅q1)q1+(x⋅q2)q2+⋯+(x⋅qn)qn.
Coordinate graph of f(x) = x² − 2 and its instantaneous tangent line at x = 1.
▲ Level 1 Foundation — Theorem 8.1: Orthogonal Sets are Linearly Independent
If S={v1,v2,…,vk} is an orthogonal set of nonzero vectors in Rn, then S is linearly independent and therefore forms an orthogonal basis for span(S).
Proof of Theorem 8.1
Suppose there exist scalars c1,c2,…,ck satisfying the linear dependence relation: c1v1+c2v2+⋯+ckvk=0. Select any index i∈{1,2,…,k} and take the inner product of both sides with vi: (j=1∑kcjvj)⋅vi=0⋅vi=0. Expanding by the bilinearity of the inner product: c1(v1⋅vi)+⋯+ci(vi⋅vi)+⋯+ck(vk⋅vi)=0. By mutual orthogonality, every term with j=i vanishes (vj⋅vi=0), reducing the equation to: ci∥vi∥2=0. Since vi=0, the squared Euclidean norm ∥vi∥2>0, forcing ci=0. Because this holds identically for every i∈{1,…,k}, the only linear combination yielding 0 is the trivial one: c1=c2=⋯=ck=0. Hence, S is linearly independent. ■
Worked Example 8.A (Level 1) — Coordinates Under Orthonormal Basis
Verify that the vectors q1=31221, q2=31−212, and q3=311−22 form an orthonormal basis for R3, and express x=900 as a linear combination of these basis vectors.
Step 1: Check orthonormality: ∥q1∥2=94+4+1=1, ∥q2∥2=94+1+4=1, ∥q3∥2=91+4+4=1. q1⋅q2=9−4+2+2=0, q1⋅q3=92−4+2=0, q2⋅q3=9−2−2+4=0. Since the 3 vectors are orthonormal in R3, they form an orthonormal basis.
Step 2: Coordinate expansion via dot products:c1=x⋅q1=900⋅2/32/31/3=6.c2=x⋅q2=900⋅−2/31/32/3=−6.c3=x⋅q3=900⋅1/3−2/32/3=3. Therefore, x=6q1−6q2+3q3. Notice that this required zero Gaussian elimination!
The Gram–Schmidt Orthogonalization Process: Given a linearly independent set of vectors {x1,x2,…,xp} in Rn, the Gram–Schmidt algorithm constructs an orthogonal basis {v1,v2,…,vp} for the same subspace via sequential orthogonal projections: v1v2v3vp=x1,=x2−v1⋅v1x2⋅v1v1,=x3−v1⋅v1x3⋅v1v1−v2⋅v2x3⋅v2v2,⋮=xp−j=1∑p−1vj⋅vjxp⋅vjvj. Normalizing yields an orthonormal basis: qi=∥vi∥vi for each i=1,…,p.
The QR Decomposition: If A is an m×n matrix with linearly independent columns, then A can be factored uniquely as: A=QR where Q is an m×n matrix whose columns form an orthonormal basis for Col(A) (satisfying QTQ=In), and R is an n×n upper triangular invertible matrix with positive diagonal entries: R=QTA=∥v1∥00⋮q1⋅x2∥v2∥0⋮q1⋅x3q2⋅x3∥v3∥⋮⋯⋯⋯⋱.
A square matrix Q∈Rn×n is called an orthogonal matrix if its columns form an orthonormal basis for Rn. The following conditions are strictly equivalent:
QTQ=QQT=In (hence Q−1=QT).
∥Qx∥=∥x∥ for all x∈Rn (isometry: preserves lengths).
(Qx)⋅(Qy)=x⋅y for all x,y∈Rn (preserves angles and inner products).
det(Q)=±1 (since det(QTQ)=(detQ)2=det(In)=1).
Worked Example 8.B (Level 2) — Gram–Schmidt & QR Factorization
Apply the Gram–Schmidt process to find the QR factorization of: A=110101.
Step 1: Orthogonalize columns x1=(1,1,0)T and x2=(1,0,1)T: First vector: v1=x1=110. Second vector: subtract projection of x2 onto v1: x2⋅v1=1(1)+0(1)+1(0)=1,∥v1∥2=12+12+02=2.v2=x2−21v1=101−1/21/20=1/2−1/21. Check orthogonality: v1⋅v2=1(1/2)+1(−1/2)+0(1)=0. ✓
● Level 3 University Rigor — Definition 8.3: Orthogonal Projections & Least Squares
Orthogonal Projection Matrix: Let W be a subspace of Rm spanned by the linearly independent columns of an m×n matrix A (where m≥n). The orthogonal projection matrixP that maps any vector b∈Rm to its closest point b^=projW(b)∈W is given by: P=A(ATA)−1AT.P satisfies two algebraic hallmarks of orthogonal projectors:
Idempotence:P2=P (projecting an already projected vector changes nothing).
Symmetry:PT=P (guaranteeing orthogonality of the projection).
The Least-Squares Problem: For an inconsistent system Ax=b, no vector x achieves Ax=b. The least-squares solutionx^ minimizes the Euclidean distance ∥Ax−b∥, which is equivalent to finding x^ such that: Ax^=projCol(A)(b).
▲ Level 3 University Rigor — Theorem 8.3: Derivation of the Normal Equations
Let A∈Rm×n and b∈Rm. A vector x^∈Rn minimizes ∥Ax−b∥2 if and only if x^ satisfies the normal equations: ATAx^=ATb. If A has full column rank (rank(A)=n), then ATA is strictly positive definite and invertible, yielding the unique least-squares solution: x^=(ATA)−1ATb.
Proof of Theorem 8.3 (Orthogonal Decomposition & Calculus)
Geometric Derivation: By the Best Approximation Theorem, the vector in Col(A) closest to b is the orthogonal projection b^=Ax^. The error vector (residual) is e=b−Ax^. By the definition of orthogonal projection, e must be orthogonal to every vector in the subspace Col(A). In particular, e is perpendicular to each column of A: aj⋅(b−Ax^)=0for each j=1,2,…,n. In matrix notation, this condition is written compactly as: AT(b−Ax^)=0⟺ATb−ATAx^=0⟺ATAx^=ATb.
Alternative Multivariable Calculus Derivation: Define the objective loss function f(x)=∥Ax−b∥2=(Ax−b)T(Ax−b)=xTATAx−2bTAx+bTb. Taking the matrix gradient with respect to x and setting it to zero: ∇xf(x)=2ATAx−2ATb=0⟹ATAx^=ATb. Since the Hessian matrix is ∇2f(x)=2ATA, which is positive semi-definite (and strictly positive definite when rank(A)=n), the stationary point x^ is a global minimum. ■
▲ Level 3 University Rigor — Theorem 8.4: The Spectral Theorem for Symmetric Matrices
Let A be an n×n real matrix. Then A is orthogonally diagonalizable (i.e., there exists an orthogonal matrix Q such that A=QDQT with D diagonal) if and only if A is symmetric (A=AT).
Consequences:
All eigenvalues of a real symmetric matrix are strictly real (never complex).
Eigenvectors corresponding to distinct eigenvalues are mutually orthogonal: λi=λj⟹vi⋅vj=0.
For every eigenvalue, its geometric multiplicity equals its algebraic multiplicity (symmetric matrices are never defective).
Worked Example 8.C (Level 3) — Least Squares via QR Decomposition
Solve the least-squares problem Ax≈b using the QR decomposition of A from Worked Example 8.B, where b=231.
Theory: When A=QR, the normal equations become: ATAx^=ATb⟺(QR)T(QR)x^=(QR)Tb⟺RT(QTQ)Rx^=RTQTb. Since QTQ=In, this simplifies to RTRx^=RTQTb. Since R is invertible, multiplying both sides by (RT)−1 yields the elegant system: Rx^=QTb. This eliminates the need to compute ATA or its matrix inverse!
Step 2: Solve upper triangular system Rx^=QTb via back-substitution:(201/23/2)(x^1x^2)=(5/21/6). From row 2: 23x^2=61⟹x^2=61⋅32=31. From row 1: 2x^1+21(31)=25⟹2x^1+31=5⟹2x^1=314⟹x^1=37. Therefore, x^=(7/31/3).
⊙ Explorer · Least squares — minimizing the sum of squared residualsinteractive
Each data point, the line's prediction, and the residual
x
y (actual)
mx + b (predicted)
residual
Drag to rotate, scroll to zoom · loads a small 3D library on first use
sum of squared residuals6.000
least-squares optimumm = 1.6, b = −0.5
The four points are (1,1), (2,3), (3,4), (4,6) — Worked example 8.B's data. The dashed vertical segments are each point's residual (actual − predicted); the readout is the sum of their squares, exactly the quantity being minimized. Drag m and b toward the stated optimum and watch the sum of squared residuals bottom out — it can't go any lower for any other line, which is exactly what solving the normal equations guarantees.
⚠️ Common Traps & University Pitfalls — Orthogonality & Least Squares
Gram–Schmidt Projection Target: When subtracting projections in Gram–Schmidt to form vk, you must project xk onto the already orthogonalized vectorsv1,…,vk−1, NOT onto the original input vectors x1,…,xk−1. Projecting onto non-orthogonal vectors destroys mutual orthogonality completely.
Order of Q and QT: For an m×n matrix Q with orthonormal columns (where m>n), QTQ=In, but QQT=Im. In fact, QQT=P is the m×m orthogonal projector onto Col(Q). Only square orthogonal matrices have QQT=I.
Singularity of ATA: The normal equations have a unique solution if and only if the columns of A are linearly independent. If A has multicollinear columns (rank <n), ATA is singular, and regularized least squares (Ridge / Tikhonov regression: (ATA+λI)x^=ATb) or pseudoinverse methods are required.
Residual Orthogonality Direction: The residual error vector e=b−Ax^ is orthogonal to the column space Col(A) (so ATe=0). It is NOT orthogonal to b. Confusing e⊥Col(A) with e⊥b is a frequent conceptual exam blunder.
Orthogonal Projection
proja(b)=a⋅aa⋅ba
Gram–Schmidt
vk=xk−∑j=1k−1∥vj∥2xk⋅vjvj
QR Least Squares
Rx^=QTb
Normal Equations
ATAx^=ATb
Orthogonal projection of y=⟨3,4⟩ onto the subspace L=span{⟨2,1⟩}. The projection y^=⟨4,2⟩ minimizes the distance to y, ensuring the error vector z=y−y^=⟨−1,2⟩ is perpendicular to L.
Practice Problem (Level 1) — Orthogonal Projection & Residual
Find the orthogonal projection of b=(4,3)T onto a=(1,2)T, and verify the residual is orthogonal to a.
Worked Example D — Euclidean Distance & Orthogonal Projection
A displacement has horizontal component Δx=5 and vertical component Δy=5. Find the squared magnitude d2 and the area of the right triangle formed with the axes.
Step-by-step Solution:
By the Pythagorean theorem, d2=(5)2+(5)2=25+25=50.
The enclosed right-triangle area is A=21(5)(5)=12.5.
Find the least-squares solution x^ to the inconsistent system Ax=b, where A=111123,b=124.
Step 2: Invert ATA and solve:det(ATA)=3(14)−6(6)=42−36=6.(ATA)−1=61(14−6−63).x^=61(14−6−63)(717)=61(98−102−42+51)=61(−49)=(−2/33/2).
[Orthogonality, Gram–Schmidt, QR Decomposition & Least Squares · Practice 2] If the governing relation for this system is g(t)=4t+4, evaluate g(6) and find the inverse input t when g(t)=32.
[Orthogonality, Gram–Schmidt, QR Decomposition & Least Squares · Practice 3] If the governing relation for this system is g(t)=5t+5, evaluate g(8) and find the inverse input t when g(t)=50.
[Orthogonality, Gram–Schmidt, QR Decomposition & Least Squares · Practice 4] If the governing relation for this system is g(t)=6t+6, evaluate g(10) and find the inverse input t when g(t)=72.
Twenty-five university-rigorous problems spanning all eight chapters — from foundational mechanics to advanced proof-based derivations and analytical applications. Try each problem before revealing the detailed worked solution.
L1
Let u=(1,−2,2,0)T and v=(2,1,0,−2)T in R4. Compute ∥u∥, ∥v∥, the inner product u⋅v, the cosine of the angle θ between them, and verify the Cauchy–Schwarz inequality.
3. Angle: cosθ=∥u∥∥v∥u⋅v=3⋅30=0⟹θ=2π rad (90∘).
4. Cauchy–Schwarz Verification: ∣u⋅v∣=0≤∥u∥∥v∥=3⋅3=9. The vectors are orthogonal, satisfying the inequality strictly. ✓
L2
Find the shortest Euclidean distance from the point P(3,−1,4) to the hyperplane H:2x−y+2z=5 in R3.
Intermediate
Hyperplane Distance Formula: For H:n⋅x=d, the perpendicular distance from x0 is: D=∥n∥∣n⋅x0−d∣. Here n=(2,−1,2)T, d=5, and x0=(3,−1,4)T.
1. Norm of normal vector:∥n∥=22+(−1)2+22=4+1+4=9=3.
2. Value at point P:n⋅x0=2(3)−1(−1)+2(4)=6+1+8=15.
3. Distance:D=3∣15−5∣=310.
L3
Prove the Parallelogram Law for arbitrary vectors u,v∈Rn: ∥u+v∥2+∥u−v∥2=2∥u∥2+2∥v∥2. What is its geometric significance?
Advanced
Algebraic Proof: Expand each squared Euclidean norm using the inner product: ∥u+v∥2=(u+v)⋅(u+v)=∥u∥2+2(u⋅v)+∥v∥2.∥u−v∥2=(u−v)⋅(u−v)=∥u∥2−2(u⋅v)+∥v∥2. Adding the two equations directly cancels the cross terms 2(u⋅v) and −2(u⋅v): ∥u+v∥2+∥u−v∥2=2∥u∥2+2∥v∥2.■
Geometric Meaning: In any parallelogram formed by vectors u and v, the sum of the squares of the lengths of the two diagonals (u+v and u−v) equals the sum of the squares of the lengths of all four sides.
L4
Express the complete solution set of the non-homogeneous linear system in parametric vector form: ⎩⎨⎧x1+2x2−x3+3x4=42x1+4x2−x3+8x4=11−x1−2x2+2x3−x4=−1
Intermediate
Step 1: Set up augmented matrix and row reduce:12−124−2−1−1238−1411−1R2−2R1,R3+R1100200−111322433R3−R2,R1+R2100200010520730.
Step 2: Identify pivot and free variables: Pivots in columns 1 and 3 (x1,x3). Free variables are x2=s, x4=t. From RREF: x1=7−2s−5t, x3=3−2t.
Perform a complete bifurcation analysis of the system in terms of parameter k∈R: ⎩⎨⎧x+y+kz=23x+4y+2z=k2x+3y−z=1 Determine all values of k for which the system has (a) a unique solution, (b) infinitely many solutions, and (c) no solution.
Advanced
Step 1: Compute determinant of coefficient matrix A:det(A)=132143k2−1. Expanding along row 1: det(A)=1(4(−1)−2(3))−1(3(−1)−2(2))+k(3(3)−4(2))=1(−10)−1(−7)+k(1)=k−3.
Step 2: Classify cases:
(a) Unique Solution: If k=3, det(A)=0; the system has full rank 3 and possesses a unique solution.
Critical Case k=3: Substitute k=3 into the augmented matrix: 13214332−1231R2−3R1,R3−2R11001113−7−72−3−3R3−R21001103−702−30. The bottom row is 0=0, so rank(A)=rank([A∣b])=2<3. Therefore, for k=3, there are infinitely many solutions (1 free variable).
(c) No Solution: There are no values of k for which the system is inconsistent.
L6
Prove that if xp is any particular solution to Ax=b, then the complete solution set is the affine subspace S={xp+xh:xh∈Null(A)}.
Intermediate
Two-part set inclusion proof:
1. Show S⊆Solution Set: Let x=xp+xh where Axp=b and Axh=0. Applying matrix multiplication: Ax=A(xp+xh)=Axp+Axh=b+0=b. Thus every vector in S is indeed a valid solution.
2. Show Solution Set⊆S: Let y be an arbitrary solution to Ax=b, so Ay=b. Define w=y−xp. Then: Aw=A(y−xp)=Ay−Axp=b−b=0. This proves w∈Null(A). Hence y=xp+w∈S. Therefore the complete solution set is precisely xp+Null(A). ■
L7
Let A and B be invertible n×n matrices. Prove algebraically that: (a) (AB)T=BTAT, and (b) (AB)−1=B−1A−1.
Intermediate
(a) Transpose Proof: The (i,j)-entry of (AB)T is the (j,i)-entry of AB: [(AB)T]ij=[AB]ji=k=1∑nAjkBki=k=1∑nBkiAjk. Using the definition of matrix transpose, Bki=[BT]ik and Ajk=[AT]kj: [(AB)T]ij=k=1∑n[BT]ik[AT]kj=[BTAT]ij. Since entries match for all i,j, (AB)T=BTAT. ■
(b) Inverse Proof: A matrix C is the inverse of AB if and only if (AB)C=In and C(AB)=In. Test C=B−1A−1: (AB)(B−1A−1)=A(BB−1)A−1=AInA−1=AA−1=In. Check reverse order: (B−1A−1)(AB)=B−1(A−1A)B=B−1InB=B−1B=In. By uniqueness of matrix inverses, (AB)−1=B−1A−1. ■
L8
Use the Gauss–Jordan elimination algorithm on the augmented block matrix [A∣I3] to compute A−1 for: A=1240−11238.
Intermediate
Step 1: Set up [A∣I]:1240−11238100010001R2−2R1,R3−4R11000−112−101−2−4010001.
Step 2: Row swap and scaling:R2↔R310001−120−11−4−2001010R3+R210001020−11−4−6001011.
Express the invertible matrix A=(1324) as a product of elementary matrices.
Advanced
Step 1: Reduce A to I2 via row operations: 1. R2→R2−3R1: Elementary matrix E1=(1−301). Result: (102−2). 2. R2→−21R2: Elementary matrix E2=(100−1/2). Result: (1021). 3. R1→R1−2R2: Elementary matrix E3=(10−21). Result: I2.
Step 2: Invert elementary matrices: We have E3E2E1A=I2⟹A=E1−1E2−1E3−1. Computing inverses of elementary row operations: E1−1=(1301),E2−1=(100−2),E3−1=(1021). Therefore: A=(1301)(100−2)(1021).
L10
Compute the determinant of the 4×4 matrix by Laplace cofactor expansion along an optimal row or column: A=201405−20−10313012.
Basic
Step 1: Choose row with maximum zeros: Row 2 has three zero entries: a21=0,a22=5,a23=0,a24=0. Expanding along Row 2: det(A)=(−1)2+2(5)214−131312=5214−131312.
Step 3: Multiply by coefficient:det(A)=5(−25)=−125.
L11
Let A=102231014. Compute det(A), find the adjugate matrix adj(A), and verify the identity Aadj(A)=det(A)I3.
Intermediate
Step 1: Compute det(A): Expanding along row 1: det(A)=1(3(4)−1(1))−2(0(4)−1(2))+0=1(11)−2(−2)=11+4=15.
Step 2: Compute Cofactor Matrix C:C11=+(12−1)=11,C12=−(0−2)=2,C13=+(0−6)=−6C21=−(8−0)=−8,C22=+(4−0)=4,C23=−(1−4)=3C31=+(2−0)=2,C32=−(1−0)=−1,C33=+(3−0)=3 The adjugate matrix is the transpose adj(A)=CT: adj(A)=112−6−8432−13.
Let M2(R) be the vector space of 2×2 real matrices. Determine, with formal verification via the Subspace Test, whether W={A∈M2(R):AT=−A} (the set of skew-symmetric matrices) is a subspace of M2(R). Find a basis and dim(W).
Intermediate
Subspace Test Verification:
Non-emptiness: The zero matrix O satisfies OT=O=−O, so O∈W.
Additivity: Let A,B∈W. Then (A+B)T=AT+BT=(−A)+(−B)=−(A+B). Thus A+B∈W.
Scalar Closure: Let A∈W and c∈R. Then (cA)T=c(AT)=c(−A)=−(cA). Thus cA∈W.
Since all 3 axioms hold, W is a vector subspace of M2(R).
General form and basis: A matrix A=(acbd) satisfies AT=−A⟺(abcd)=(−a−c−b−d). This requires a=−a⟹a=0, d=−d⟹d=0, and c=−b. Thus: A=(0−bb0)=b(0−110). A basis for W is {(0−110)}, and dim(W)=1.
L14
For the matrix A=123246011145, find bases for Row(A), Col(A), and Null(A). Verify the Rank–Nullity Theorem.
Intermediate
Step 1: Row reduction to RREF:123246011145R2−2R1,R3−3R1100200011122R3−R2100200010120.
Step 2: Subspace Bases:
Basis for Row(A): Non-zero rows of RREF: {(1,2,0,1),(0,0,1,2)}.
Basis for Col(A): Pivot columns of original matrix A (columns 1 and 3): ⎩⎨⎧123,011⎭⎬⎫.
Basis for Null(A): Free variables x2=s,x4=t. From RREF: x1=−2s−t, x3=−2t. x=s−2100+t−10−21⟹Basis=⎩⎨⎧−2100,−10−21⎭⎬⎫.
Verification:rank(A)=2, nullity(A)=2. Total columns n=4. rank(A)+nullity(A)=2+2=4=n. ✓
L15
Let B={1,1+x,1+x+x2} be an ordered basis for the polynomial space P2. Find the coordinate vector [p(x)]B for p(x)=3−2x+5x2.
Basic
Set up the coordinate equation:p(x)=c1(1)+c2(1+x)+c3(1+x+x2)=(c1+c2+c3)+(c2+c3)x+c3x2. Equating coefficients of like powers with 3−2x+5x2:
Coefficient of x2: c3=5
Coefficient of x: c2+c3=−2⟹c2+5=−2⟹c2=−7
Constant term: c1+c2+c3=3⟹c1−7+5=3⟹c1=5
Therefore, the coordinate vector is: [p(x)]B=5−75.
L16
Prove that if λ is an eigenvalue of an invertible matrix A, then λ=0 and λ−1 is an eigenvalue of A−1 with the exact same corresponding eigenvector.
Intermediate
Step 1: Show λ=0: By definition of an eigenvalue, there exists a nonzero vector v=0 such that Av=λv. If λ=0, then Av=0, which means v∈Null(A), contradicting the assumption that A is invertible. Thus λ=0.
Step 2: Show λ−1 is an eigenvalue of A−1: Multiply both sides of Av=λv by A−1 from the left: A−1(Av)=A−1(λv)⟹(A−1A)v=λ(A−1v)⟹v=λ(A−1v). Dividing both sides by the nonzero scalar λ: A−1v=λ1v=λ−1v. Since v=0, this equation precisely states that λ−1 is an eigenvalue of A−1 with the identical eigenvector v. ■
L17
Find all eigenvalues, their algebraic and geometric multiplicities, and a basis for each eigenspace for: A=300130002. Is A diagonalizable?
Intermediate
Step 1: Characteristic equation: Since A is upper triangular, its eigenvalues are the diagonal entries: det(A−λI)=(3−λ)2(2−λ)=0. Distinct eigenvalues: λ1=3 (algebraic multiplicity α1=2), and λ2=2 (algebraic multiplicity α2=1).
Step 2: Eigenspace for λ=3:(A−3I)v=00010000−1v1v2v3=000. We obtain v2=0 and −v3=0⟹v3=0, while v1 is free. Thus: E3=span⎩⎨⎧100⎭⎬⎫⟹geometric multiplicity γ1=1.
Step 3: Eigenspace for λ=2:(A−2I)v=100110000v1v2v3=000⟹v1=0,v2=0,v3 free⟹E2=span⎩⎨⎧001⎭⎬⎫.
Conclusion: For λ=3, γ1=1<α1=2. Because the geometric multiplicity falls strictly below the algebraic multiplicity, A is defective and NOT diagonalizable.
L18
Diagonalize the matrix A=(1221) by finding invertible P and diagonal D such that A=PDP−1. Then compute Ak for any positive integer k.
Advanced
Step 1: Eigenvalues of A:det(A−λI)=(1−λ)2−4=λ2−2λ−3=(λ−3)(λ+1)=0⟹λ1=3,λ2=−1.
Step 2: Eigenvectors:
For λ1=3: (−222−2)v1=0⟹v1=(11).
For λ2=−1: (2222)v2=0⟹v2=(−11).
Step 3: Construct P, D, and P−1:P=(11−11),D=(300−1),P−1=21(1−111).
Solve the initial value problem for the coupled system of linear differential equations: x′(t)=(1243)x(t),x(0)=(51).
Advanced
Step 1: Eigenvalues of coefficient matrix:det(A−λI)=(1−λ)(3−λ)−8=λ2−4λ−5=(λ−5)(λ+1)=0⟹λ1=5,λ2=−1.
Step 2: Eigenvectors:
For λ1=5: (A−5I)=(−424−2)⟹v1=(11).
For λ2=−1: (A+I)=(2244)⟹v2=(−21).
Step 3: General solution and initial conditions:x(t)=c1e5t(11)+c2e−t(−21). At t=0: c1(11)+c2(−21)=(51)⟺(11−21)(c1c2)=(51). Subtracting rows: 3c2=−4⟹c2=−4/3; then c1=5+2(−4/3)=7/3. Therefore: x(t)=37e5t(11)−34e−t(−21)=(37e5t+38e−t37e5t−34e−t).
L20
Verify the Cayley–Hamilton Theorem for A=(2314), and use it to compute A−1 without using cofactor formulas or row reduction.
Step 3: Solve for A−1: Multiply the operator identity A2−6A+5I2=O by A−1: A−6I2+5A−1=O⟹5A−1=6I2−A⟹A−1=51(6I2−A). Evaluating directly: A−1=51((6006)−(2314))=51(4−3−12).
L21
Find the standard matrix [T] of the linear operator T:R2→R2 that performs a horizontal shear by factor 2, followed by a counterclockwise rotation of 45∘.
Basic
Step 1: Standard matrix of shear [S]:S(x,y)=(x+2y,y)⟹[S]=(1021).
Step 2: Standard matrix of rotation [R]:[R]=(cos(45∘)sin(45∘)−sin(45∘)cos(45∘))=22(11−11).
im(T): Pivot columns 1 and 2 of A. Basis = ⎩⎨⎧123,257⎭⎬⎫, rank = 2.
Isomorphism check: Since ker(T)={0}, T is not injective (and rank(T)=2<3 means not surjective). Therefore, T is not an isomorphism.
L23
Let T:R2→R2 have standard matrix A=(4213). Find the matrix representation [T]B relative to the basis B={(11),(−12)} via similarity transformation P−1AP.
Step 3: Compute [T]B=P−1(AP):[T]B=31(2−111)(55−24)=31(10+5−5+5−4+42+4)=31(15006)=(5002). (Notice that B is an eigenbasis of A, completely diagonalizing T!)
L24
Apply the Gram–Schmidt orthogonalization process to the columns of A to find its QR factorization: A=12−2212.
Intermediate
Step 1: Orthogonalize x1=(1,2,−2)T and x2=(2,1,2)T: First vector: v1=x1=12−2. Norm: ∥v1∥=12+22+(−2)2=9=3. Subtract projection of x2 onto v1: x2⋅v1=2(1)+1(2)+2(−2)=2+2−4=0. Since x2⋅v1=0, the columns were already orthogonal! Thus v2=x2=212. Norm: ∥v2∥=22+12+22=9=3.
Step 2: Normalize to construct Q:q1=3112−2,q2=31212⟹Q=1/32/3−2/32/31/32/3.
Find the least-squares regression line y=mx+b for the points (0,1),(1,3),(2,2),(3,6) using the normal equations ATAx^=ATy. Compute the residual vector e=y−Ax^ and verify that e⊥Col(A).
Advanced
Step 1: Set up matrix equation Ax=y:A=01231111,y=1326,x=(mb).
Step 2: Form normal equations:ATA=(01112131)01231111=(0+1+4+90+1+2+30+1+2+31+1+1+1)=(14664).ATy=(01112131)1326=(0+3+4+181+3+2+6)=(2512).
Step 3: Invert and solve:det(ATA)=14(4)−6(6)=56−36=20.x^=201(4−6−614)(2512)=201(100−72−150+168)=201(2818)=(1.40.9). Thus the least-squares line is y=1.4x+0.9.
Step 4: Residual and Orthogonality Check:Ax^=0(1.4)+0.91(1.4)+0.92(1.4)+0.93(1.4)+0.9=0.92.33.75.1⟹e=y−Ax^=1−0.93−2.32−3.76−5.1=0.10.7−1.70.9. Testing orthogonality with columns of A: a1⋅e=0(0.1)+1(0.7)+2(−1.7)+3(0.9)=0+0.7−3.4+2.7=0. ✓ a2⋅e=1(0.1)+1(0.7)+1(−1.7)+1(0.9)=0.1+0.7−1.7+0.9=0. ✓ The residual error vector is strictly orthogonal to Col(A), confirming the geometric optimality of the solution.
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Last updated October 16, 2026
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