01.01.E1 · foundations / linear-algebra

Linear algebra exercise pack (Apostol Vol. 2 Ch. 1-5 supplement)

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Formal definition of the pack Intermediate

Apostol Vol. 2 Chapters 1-5 build the linear-algebra spine: abstract linear spaces with subspace, span, independence, and dimension (Ch. 1); linear transformations, the rank-nullity theorem, and the matrix of a map in fixed bases (Ch. 2); the determinant as the unique multilinear alternating normalised function, with expansion and product rules (Ch. 3); eigenvalues, eigenvectors, the characteristic polynomial, and diagonalisability (Ch. 4); and the inner-product theory of Euclidean spaces — Gram-Schmidt, orthogonal projection, and the spectral theorem for symmetric operators (Ch. 5).

This pack collects ten problems drawn from those chapters — three easy, four medium, three hard — each with a hint and a complete worked solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The problems test operational competence: computing a dimension, applying rank-nullity, evaluating a determinant by structure rather than brute force, diagonalising a matrix, and orthonormalising a basis. Several deliberately reward a structural argument over a computational one, in Apostol's style.

Conventions follow the prerequisite units: $V$ , $W$ denote finite-dimensional vector spaces over a field $F$ ( $R$ or $C$ unless stated); $T : V \to W$ a linear map; $ker T$ and $im T$ its kernel and image; $χ_{A} (λ) = det (λ I - A)$ the characteristic polynomial; $⟨ \cdot, \cdot ⟩$ an inner product.

Key theorem with full solution Intermediate

Before the pack proper, we work one problem in full as an exemplar of the format. The remaining nine follow the same structure (problem, hint, full answer in <details> blocks).

Lead problem. Rank-nullity and the dimension of a sum. Let $U, W$ be subspaces of a finite-dimensional space $V$ . Prove $dim (U + W) + dim (U \cap W) = dim U + dim W$ , and deduce that if $dim U + dim W > dim V$ then $U \cap W \neq = {0}$ .

Solution. Let ${e_{1}, \dots, e_{k}}$ be a basis of $U \cap W$ , so $dim (U \cap W) = k$ . Since $U \cap W \subseteq U$ , extend it to a basis ${e_{1}, \dots, e_{k}, u_{1}, \dots, u_{p}}$ of $U$ , so $dim U = k + p$ . Likewise extend to a basis ${e_{1}, \dots, e_{k}, w_{1}, \dots, w_{q}}$ of $W$ , so $dim W = k + q$ .

Claim: $B = {e_{1}, \dots, e_{k}, u_{1}, \dots, u_{p}, w_{1}, \dots, w_{q}}$ is a basis of $U + W$ . It spans, because any element of $U + W$ is a sum of a $U$ -vector and a $W$ -vector, each expressible in the listed vectors. For independence, suppose $\sum a_{i} e_{i} + \sum b_{j} u_{j} + \sum c_{l} w_{l} = 0$ . Then $\sum c_{l} w_{l} = - \sum a_{i} e_{i} - \sum b_{j} u_{j}$ lies in $U$ (right side) and in $W$ (left side), so it lies in $U \cap W$ , hence is a combination of the $e_{i}$ . But the $w_{l}$ together with the $e_{i}$ form a basis of $W$ , so the only way $\sum c_{l} w_{l}$ is a combination of the $e_{i}$ is $c_{l} = 0$ for all $l$ . The relation reduces to $\sum a_{i} e_{i} + \sum b_{j} u_{j} = 0$ , a relation among a basis of $U$ , forcing all $a_{i}, b_{j} = 0$ . So $B$ is independent.

Therefore $dim (U + W) = k + p + q = (k + p) + (k + q) - k = dim U + dim W - dim (U \cap W)$ , which rearranges to the stated identity.

For the deduction: $dim (U \cap W) = dim U + dim W - dim (U + W) \geq dim U + dim W - dim V > 0$ , using $U + W \subseteq V$ . A space of positive dimension contains a nonzero vector, so $U \cap W \neq = {0}$ . $□$

This is the Grassmann dimension formula (Apostol Ch. 1 §1.13). The deduction is the standard pigeonhole on dimensions: two subspaces whose dimensions overcrowd the ambient space must meet nontrivially.

Exercises Intermediate

Exercise 3 (easy, symbolic). Eigenvalues of a triangular matrix.

Show that the eigenvalues of an upper-triangular matrix are exactly its diagonal entries.

Hint

The determinant of an upper-triangular matrix is the product of its diagonal entries. Apply this to $λ I - A$ .

Answer

Let $A = (a_{ij})$ be upper-triangular, so $a_{ij} = 0$ for $i > j$ . Then $λ I - A$ is also upper-triangular, with diagonal entries $λ - a_{ii}$ . The determinant of a triangular matrix is the product of its diagonal entries, so $$ \chi_A(\lambda) = \det(\lambda I - A) = \prod_{i=1}^n (\lambda - a_{ii}). $$ The roots of $χ_{A}$ are the eigenvalues, and they are precisely $a_{11}, \dots, a_{nn}$ (with algebraic multiplicity equal to the number of times each value appears on the diagonal). The same argument applies to lower-triangular matrices.

Exercise 4 (medium, symbolic). Diagonalise a $2 \times 2$ matrix.

Diagonalise $$ A = \begin{pmatrix} 4 & 1 \ 2 & 3 \end{pmatrix}: $$ find an invertible $P$ and diagonal $D$ with $A = P D P^{- 1}$ .

Hint

Find the eigenvalues from $det (λ I - A) = 0$ , then solve $(A - λ I) v = 0$ for each eigenvector. The columns of $P$ are the eigenvectors; $D$ holds the eigenvalues.

Answer

Characteristic polynomial: $χ_{A} (λ) = (λ - 4) (λ - 3) - 2 = λ^{2} - 7 λ + 10 = (λ - 5) (λ - 2)$ . Eigenvalues $λ = 5, 2$ .

For $λ = 5$ : $(A - 5 I) = (- 1 2 1 - 2)$ , kernel spanned by $v_{1} = (1, 1)$ .

For $λ = 2$ : $(A - 2 I) = (2211)$ , kernel spanned by $v_{2} = (1, - 2)$ .

Then $$ P = \begin{pmatrix} 1 & 1 \ 1 & -2 \end{pmatrix}, \qquad D = \begin{pmatrix} 5 & 0 \ 0 & 2 \end{pmatrix}, \qquad A = P D P^{-1}. $$ Check: $det P = - 3 \neq = 0$ , so $P$ is invertible and the diagonalisation is valid. Two distinct eigenvalues in a $2$ -dimensional space always force diagonalisability.

Exercise 6 (medium, proof). Eigenvectors for distinct eigenvalues are independent.

Let $T : V \to V$ be linear with eigenvectors $v_{1}, \dots, v_{k}$ for pairwise distinct eigenvalues $λ_{1}, \dots, λ_{k}$ . Prove $v_{1}, \dots, v_{k}$ are linearly independent.

Hint

Induct on $k$ . Apply $T$ to a dependence relation, and also multiply the relation by $λ_{k}$ ; subtract to eliminate $v_{k}$ and invoke the inductive hypothesis.

Answer

Induct on $k$ . For $k = 1$ , a single eigenvector is nonzero by definition, hence independent.

Assume the result for $k - 1$ . Suppose $\sum_{i = 1}^{k} c_{i} v_{i} = 0$ . Apply $T$ : since $T v_{i} = λ_{i} v_{i}$ , we get $\sum_{i = 1}^{k} c_{i} λ_{i} v_{i} = 0$ . Multiply the original relation by $λ_{k}$ : $\sum_{i = 1}^{k} c_{i} λ_{k} v_{i} = 0$ . Subtract: $$ \sum_{i=1}^{k-1} c_i (\lambda_i - \lambda_k) v_i = 0 $$ (the $i = k$ term cancels). By the inductive hypothesis, $v_{1}, \dots, v_{k - 1}$ are independent, so each coefficient $c_{i} (λ_{i} - λ_{k}) = 0$ . Since the eigenvalues are distinct, $λ_{i} - λ_{k} \neq = 0$ for $i < k$ , forcing $c_{i} = 0$ for $i = 1, \dots, k - 1$ . The original relation collapses to $c_{k} v_{k} = 0$ , and $v_{k} \neq = 0$ gives $c_{k} = 0$ . So all coefficients vanish.

Corollary: a map with $n = dim V$ distinct eigenvalues is diagonalisable.

Exercise 7 (medium, numeric). Gram-Schmidt in $R^{3}$ .

Apply Gram-Schmidt to $a_{1} = (1, 1, 0)$ , $a_{2} = (1, 0, 1)$ , $a_{3} = (0, 1, 1)$ with the standard inner product to produce an orthonormal basis.

Hint

Set $u_{1} = a_{1}$ , then subtract the projection of $a_{2}$ onto $u_{1}$ , then subtract the projections of $a_{3}$ onto $u_{1}$ and $u_{2}$ . Normalise at the end.

Answer

$u_{1} = a_{1} = (1, 1, 0)$ , $∥ u_{1} ∥^{2} = 2$ .

$u_{2} = a_{2} - \frac{⟨ a _{2} , u _{1} ⟩}{⟨ u _{1} , u _{1} ⟩} u_{1} = (1, 0, 1) - \frac{1}{2} (1, 1, 0) = (\frac{1}{2}, - \frac{1}{2}, 1)$ , with $∥ u_{2} ∥^{2} = \frac{1}{4} + \frac{1}{4} + 1 = \frac{3}{2}$ .

$u_{3} = a_{3} - \frac{⟨ a _{3} , u _{1} ⟩}{∥ u _{1} ∥ ^{2}} u_{1} - \frac{⟨ a _{3} , u _{2} ⟩}{∥ u _{2} ∥ ^{2}} u_{2}$ . Here $⟨ a_{3}, u_{1} ⟩ = 1$ and $⟨ a_{3}, u_{2} ⟩ = - \frac{1}{2} + 1 = \frac{1}{2}$ , so $$ u_3 = (0,1,1) - \tfrac{1}{2}(1,1,0) - \frac{1/2}{3/2}(\tfrac12,-\tfrac12,1) = (0,1,1) - (\tfrac12,\tfrac12,0) - (\tfrac16,-\tfrac16,\tfrac13) = (-\tfrac{2}{3}, \tfrac{2}{3}, \tfrac{2}{3}). $$ Normalising: $e_{1} = \frac{1}{2} (1, 1, 0)$ , $e_{2} = \frac{2}{3} (\frac{1}{2}, - \frac{1}{2}, 1) = \frac{1}{6} (1, - 1, 2)$ , $e_{3} = \frac{1}{3} (- 1, 1, 1)$ . One checks $⟨ e_{i}, e_{j} ⟩ = δ_{ij}$ .

Exercise 8 (hard, proof). Vandermonde determinant.

Prove that for scalars $x_{1}, \dots, x_{n}$ , $$ \det\begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \ \vdots & & & & \vdots \ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \end{pmatrix} = \prod_{1 \leq i < j \leq n} (x_j - x_i). $$

Hint

Induct on $n$ . Use column operations $C_{k} \mapsto C_{k} - x_{1} C_{k - 1}$ (from the right) to introduce a row of zeros except the first entry, then factor $(x_{i} - x_{1})$ from each remaining row.

Answer

Let $V_{n}$ be the determinant. For $n = 1$ , $V_{1} = 1$ (empty product). For $n = 2$ , $V_{2} = x_{2} - x_{1}$ , matching.

Inductive step: perform the column operations $C_{k} \mapsto C_{k} - x_{1} C_{k - 1}$ for $k = n, n - 1, \dots, 2$ (in that order). The first row becomes $(1, 0, 0, \dots, 0)$ . Row $i$ becomes $(1, x_{i} - x_{1}, x_{i}^{2} - x_{1} x_{i}, \dots) = (1, x_{i} - x_{1}, x_{i} (x_{i} - x_{1}), \dots, x_{i}^{n - 2} (x_{i} - x_{1}))$ . These operations leave the determinant unchanged.

Expand along the first row (only the $(1, 1)$ entry survives): $V_{n}$ equals the $(n - 1) \times (n - 1)$ determinant of the matrix with rows indexed $i = 2, \dots, n$ and entries $x_{i}^{k} (x_{i} - x_{1})$ for $k = 0, \dots, n - 2$ . Factor $(x_{i} - x_{1})$ out of row $i$ : $$ V_n = \prod_{i=2}^n (x_i - x_1) \cdot \det(x_i^{,k}){i=2,\ldots,n;, k=0,\ldots,n-2} = \prod{i=2}^n (x_i - x_1) \cdot V_{n-1}(x_2, \ldots, x_n). $$ By the inductive hypothesis $V_{n - 1} = \prod_{2 \leq i < j \leq n} (x_{j} - x_{i})$ . Combining gives $V_{n} = \prod_{1 \leq i < j \leq n} (x_{j} - x_{i})$ . The determinant is nonzero iff the $x_{i}$ are distinct — the basis for Lagrange interpolation.

Exercise 9 (hard, proof). Symmetric matrices have real eigenvalues.

Let $A$ be a real symmetric $n \times n$ matrix. Prove every eigenvalue of $A$ is real, and that eigenvectors for distinct eigenvalues are orthogonal.

Hint

Work over $C$ with the Hermitian inner product. Compare $⟨ A v, v ⟩$ and $⟨ v, A v ⟩$ using $A = A^{T}$ real. For orthogonality, compute $⟨ A v_{1}, v_{2} ⟩$ two ways.

Answer

Real eigenvalues. Let $λ \in C$ be an eigenvalue with eigenvector $v \in C^{n} ∖ {0}$ , so $A v = λ v$ . Use the Hermitian inner product $⟨ x, y ⟩ = \sum \overline{x_{i}} y_{i}$ . Then $$ \lambda \langle v, v\rangle = \langle v, Av\rangle = \langle A^* v, v\rangle = \langle A^T v, v \rangle^{}!!, $$ and since $A$ is real symmetric, $A^ = \overline{A}^T = A $. T h u s$ \langle v, Av\rangle = \langle Av, v\rangle = \overline{\lambda}\langle v, v\rangle $. E q u a t in g,$ \lambda \langle v,v\rangle = \overline{\lambda}\langle v,v\rangle $, an d$ \langle v,v\rangle > 0 $f or ces$ \lambda = \overline{\lambda} $, so$ \lambda \in \mathbb{R}$.

Orthogonality. Let $A v_{1} = λ_{1} v_{1}$ , $A v_{2} = λ_{2} v_{2}$ with $λ_{1} \neq = λ_{2}$ (both real). Then $$ \lambda_1 \langle v_1, v_2\rangle = \langle Av_1, v_2\rangle = \langle v_1, A v_2\rangle = \lambda_2 \langle v_1, v_2\rangle, $$ using $A = A^{T}$ to move $A$ across the real inner product. Hence $(λ_{1} - λ_{2}) ⟨ v_{1}, v_{2} ⟩ = 0$ , and $λ_{1} \neq = λ_{2}$ gives $⟨ v_{1}, v_{2} ⟩ = 0$ . This is the heart of the spectral theorem (Apostol Ch. 5): a real symmetric matrix has an orthonormal eigenbasis.

Exercise 10 (hard, proof). Cayley-Hamilton for a $2 \times 2$ matrix, and the trace-determinant form.

For a $2 \times 2$ matrix $A$ , prove the Cayley-Hamilton theorem $A^{2} - (tr A) A + (det A) I = 0$ by direct computation, and use it to express $A^{- 1}$ in terms of $A$ when $A$ is invertible.

Hint

The characteristic polynomial of a $2 \times 2$ matrix is $λ^{2} - (tr A) λ + det A$ . Substitute $A$ for $λ$ and verify entrywise. For the inverse, multiply the identity by $A^{- 1}$ .

Answer

Write $A = (a c b d)$ , so $tr A = a + d$ , $det A = a d - b c$ . Compute $$ A^2 = \begin{pmatrix} a^2 + bc & ab + bd \ ca + dc & cb + d^2\end{pmatrix}. $$ Then $$ A^2 - (a+d)A + (ad - bc)I = \begin{pmatrix} a^2 + bc - (a+d)a + ad - bc & ab + bd - (a+d)b \ ca + dc - (a+d)c & cb + d^2 - (a+d)d + ad - bc\end{pmatrix}. $$ The $(1, 1)$ entry is $a^{2} + b c - a^{2} - d a + a d - b c = 0$ ; the $(1, 2)$ entry is $ab + b d - ab - d b = 0$ ; similarly $(2, 1)$ and $(2, 2)$ vanish. So $A^{2} - (tr A) A + (det A) I = 0$ .

If $A$ is invertible ( $det A \neq = 0$ ), multiply by $A^{- 1}$ : $A - (tr A) I + (det A) A^{- 1} = 0$ , so $$ A^{-1} = \frac{1}{\det A}\big((\operatorname{tr} A)I - A\big) = \frac{1}{ad-bc}\begin{pmatrix} d & -b \ -c & a\end{pmatrix}, $$ recovering the standard $2 \times 2$ inverse formula from Cayley-Hamilton (Apostol Ch. 4 §4.16).

Exercise pack. Apostol Vol. 2 Chapters 1-5 supplement: linear spaces and dimension, linear transformations and rank-nullity, determinants, eigenvalues, and the inner-product / spectral theory of Euclidean spaces.

Prerequisites

01.01.04 pending
01.01.05 pending
01.01.07 pending
01.01.08 pending
01.01.09 pending

Tier anchors

beginner
intermediate: Apostol Vol. 2 exercises (Ch. 1 §1.10-1.17, Ch. 2 §2.4-2.21, Ch. 3 §3.6-3.17, Ch. 4 §4.4-4.21, Ch. 5 §5.5-5.20)
master

References

Apostol — Calculus, Volume II: Multi-Variable Calculus and Linear Algebra · Ch. 1-5 exercises: linear spaces and dimension (Ch. 1), linear transformations and matrices (Ch. 2), determinants (Ch. 3), eigenvalues and eigenvectors (Ch. 4), eigenvalues of operators on Euclidean spaces (Ch. 5)
Halmos — Finite-Dimensional Vector Spaces · §§1-79 — abstract linear-algebra problems on dimension, linear maps, determinants, spectral theory
Hoffman-Kunze — Linear Algebra · Ch. 3-9 — matrix algebra, determinants, characteristic values, inner-product spaces

Estimated time

intermediate: 5h