Eigenvalues and eigenvectors — characteristic polynomial: Exercises

Before we start, one shared picture. An eigenvector is a direction the matrix does not turn — it only stretches it. The figure shows a generic vector getting rotated and stretched (orange) versus an eigenvector that stays on its own line (green).

Figure — Eigenvalues and eigenvectors — characteristic polynomial

Level 1 — Recognition

Exercise L1.1

For $A = (4007)$ , is $x = (10)$ an eigenvector? If so, with what eigenvalue?

Recall Solution

Compute $A x$ and compare with a scalar multiple of $x$ . $A x = (4007) (10) = (40) = 4 (10) .$ WHAT we did: applied the matrix. WHY: the definition $A x = λ x$ is a test — plug the vector in and see if the output is a multiple of the input. RESULT: output is $4$ times the input, same direction, so yes — eigenvector with $== λ = 4 ==$ .

Exercise L1.2

Write the characteristic polynomial $p (λ) = det (A - λ I)$ symbolically for a general $2 \times 2$ matrix $A = (a c b d)$ . State its degree.

Recall Solution

Subtract $λ$ from each diagonal entry (that is what $A - λ I$ means: $I$ has $1$ 's on the diagonal, so $λ I$ has $λ$ 's there): $A - λ I = (a - λ c b d - λ) .$ Take the determinant (main-diagonal product minus off-diagonal product): $p (λ) = (a - λ) (d - λ) - b c = λ^{2} - (a + d) λ + (a d - b c) .$ Degree $= 2$ , matching the size of the matrix (an $n \times n$ matrix gives degree $n$ ).

Level 2 — Application

Exercise L2.1

Find both eigenvalues of $A = (2123)$ and verify them with the trace/determinant check.

Recall Solution

$p (λ) = (2 - λ) (3 - λ) - (2) (1) = λ^{2} - 5 λ + 6 - 2 = λ^{2} - 5 λ + 4.$ WHY the trace/det shortcut: $p (λ) = λ^{2} - tr (A) λ + det (A)$ , so we expect coefficient $- tr (A) = - 5$ and constant $det (A) = 4$ . Factor: $λ^{2} - 5 λ + 4 = (λ - 1) (λ - 4) \Rightarrow λ_{1} = 1, λ_{2} = 4.$ Check (see Trace of a Matrix and Determinants): $λ_{1} + λ_{2} = 5 = tr (A)$ ✓, $λ_{1} λ_{2} = 4 = det (A)$ ✓.

Exercise L2.2

Find an eigenvector of $A = (2123)$ for $λ = 4$ .

Recall Solution

Solve $(A - 4 I) x = 0$ — the eigenvector lives in the null space of $A - λ I$ . $A - 4 I = (- 2 1 2 - 1) .$ Row 1 says $- 2 x_{1} + 2 x_{2} = 0 \Rightarrow x_{1} = x_{2}$ . (Row 2 says the same: $x_{1} - x_{2} = 0$ .) One free direction: $x = (11) .$ Verify: $A (1 1) = (1 + 3 2 + 2) = (4 4) = 4 (1 1)$ ✓.

Exercise L2.3

Find the eigenvalues of the triangular matrix $A = 500 9 - 1 0 - 3 42$ .

Recall Solution

WHY no work is needed: for a triangular matrix $det (A - λ I) = \prod_{i} (a_{ii} - λ)$ , since the determinant of a triangular matrix is the product of its diagonal entries. So $p (λ) = (5 - λ) (- 1 - λ) (2 - λ) = 0 \Rightarrow λ = 5, - 1, 2.$ The eigenvalues are exactly the diagonal entries $== 5, - 1, 2 ==$ .

Level 3 — Analysis

Exercise L3.1

For $A = (4014)$ , find the eigenvalue, its algebraic multiplicity, and its geometric multiplicity. Is $A$ defective?

Recall Solution

Triangular, so $p (λ) = (4 - λ)^{2}$ . Root $λ = 4$ appears twice → algebraic multiplicity $= 2$ . Geometric multiplicity = number of independent eigenvectors = dimension of the null space of $A - 4 I$ : $A - 4 I = (0010) .$ This forces $x_{2} = 0$ , with $x_{1}$ free: only one direction $(0 1)$ . So geometric multiplicity $= 1$ . Since geometric $(1) <$ algebraic $(2)$ , the matrix is $== defective ==$ (not diagonalizable — see Diagonalization).

Exercise L3.2

The rotation matrix $R = (01 - 1 0)$ turns every real vector by $9 0^{\circ}$ . Explain via the characteristic polynomial why it has no real eigenvectors, and find its complex eigenvalues.

Recall Solution

$p (λ) = det (- λ 1 - 1 - λ) = (- λ) (- λ) - (- 1) (1) = λ^{2} + 1.$ WHY no real roots: $λ^{2} + 1 = 0$ means $λ^{2} = - 1$ , and no real number squares to a negative. Geometrically this matches the picture — a $9 0^{\circ}$ rotation turns every nonzero vector off its own line, so no real direction is preserved. Using Complex Numbers: $λ = \pm i .$ Check: sum $= i + (- i) = 0 = tr (R)$ ✓, product $= i \cdot (- i) = - i^{2} = 1 = det (R)$ ✓.

Exercise L3.3

Find all eigenvalues of $A = (1221)$ and its two eigenvectors. Note that the eigenvectors are perpendicular — why is that expected here?

Recall Solution

$p (λ) = (1 - λ)^{2} - 4 = λ^{2} - 2 λ - 3 = (λ - 3) (λ + 1) \Rightarrow λ = 3, - 1.$ For $λ = 3$ : $A - 3 I = (- 2 2 2 - 2)$ , so $- x_{1} + x_{2} = 0$ , giving $x = (1 1)$ . For $λ = - 1$ : $A + I = (2222)$ , so $x_{1} + x_{2} = 0$ , giving $x = (- 1 1)$ . Dot product $(1 1) \cdot (- 1 1) = 1 - 1 = 0$ → perpendicular. WHY expected: $A$ is symmetric ( $A = A^{⊤}$ ), and symmetric matrices always have perpendicular eigenvectors for distinct eigenvalues.

Level 4 — Synthesis

Exercise L4.1

A $2 \times 2$ matrix has $tr (A) = 6$ and $det (A) = 8$ . Without knowing $A$ , find its eigenvalues.

Recall Solution

WHY this works without $A$ : the characteristic polynomial of a $2 \times 2$ matrix depends only on trace and determinant: $p (λ) = λ^{2} - tr (A) λ + det (A)$ . $λ^{2} - 6 λ + 8 = 0 \Rightarrow (λ - 2) (λ - 4) = 0 \Rightarrow λ = 2, 4.$ Check: $2 + 4 = 6$ ✓, $2 \cdot 4 = 8$ ✓.

Exercise L4.2

Given $λ = 2$ is an eigenvalue of $A = (31 k 2)$ , find $k$ .

Recall Solution

$λ = 2$ is a root means $p (2) = det (A - 2 I) = 0$ : $A - 2 I = (11 k 0), det = (1) (0) - (k) (1) = - k .$ Set $- k = 0 \Rightarrow k = 0$ . Check: with $k = 0$ , $A = (3102)$ is triangular with diagonal $3, 2$ — eigenvalues $3$ and $2$ , and indeed $2$ is one of them ✓.

Exercise L4.3

Let $A = (2112)$ (from the parent note, eigenvalues $1$ and $3$ ). Show directly that $A^{2}$ has eigenvalues $1$ and $9$ , illustrating the rule "if $A x = λ x$ then $A^{2} x = λ^{2} x$ ."

Recall Solution

WHY the rule holds: apply $A$ twice — $A^{2} x = A (A x) = A (λ x) = λ (A x) = λ^{2} x$ . Same eigenvector, eigenvalue squared. Direct confirmation: $A^{2} = (2112)^{2} = (5445)$ . $p_{A^{2}} (μ) = (5 - μ)^{2} - 16 = μ^{2} - 10 μ + 9 = (μ - 1) (μ - 9) \Rightarrow μ = 1, 9.$ And indeed $1 = 1^{2}$ , $9 = 3^{2}$ ✓ — the squares of $A$ 's eigenvalues.

Level 5 — Mastery

Exercise L5.1

Prove that a matrix $A$ is singular (has $det A = 0$ ) if and only if $0$ is one of its eigenvalues.

Recall Solution

WHAT " $0$ is an eigenvalue" means: $p (0) = det (A - 0 \cdot I) = det (A)$ . So $0$ is a root of the characteristic polynomial exactly when $det (A) = 0$ . $0 is an eigenvalue ⟺ p (0) = 0 ⟺ det (A) = 0 ⟺ A singular .$ Geometric meaning: $A x = 0 \cdot x = 0$ for some nonzero $x$ means $A$ collapses that direction to zero — a nontrivial null space, which is precisely singularity. $■$

Exercise L5.2

For $A = (2112)$ , verify the Cayley–Hamilton theorem: $A$ satisfies its own characteristic polynomial, i.e. $p (A) = A^{2} - 4 A + 3 I = 0$ .

Recall Solution

From the parent note, $p (λ) = λ^{2} - 4 λ + 3$ . The Cayley–Hamilton Theorem says substituting the matrix for $λ$ (and $I$ for the constant term) gives the zero matrix. $A^{2} = (5445), 4 A = (8448), 3 I = (3003) .$ $A^{2} - 4 A + 3 I = (5 - 8 + 3 4 - 4 + 0 4 - 4 + 0 5 - 8 + 3) = (0000) ✓ .$

Exercise L5.3

Find the characteristic polynomial and eigenvalues of the $3 \times 3$ matrix $A = 214035 00 - 1$ , and state the sum and product of the eigenvalues, confirming them against trace and $det$ .

Recall Solution

$A$ is lower-triangular, so eigenvalues are the diagonal entries directly: $p (λ) = (2 - λ) (3 - λ) (- 1 - λ) \Rightarrow λ = 2, 3, - 1.$ Sum $= 2 + 3 - 1 = 4 = tr (A)$ (the general fact "sum of eigenvalues $=$ trace" holds for all $n$ ). Product $= 2 \cdot 3 \cdot (- 1) = - 6 = det (A)$ (also triangular = product of diagonal). Both structural facts confirmed ✓.

Recall wrap-up

Recall Quick self-quiz

To find eigenvalues you solve ::: $det (A - λ I) = 0$ To find eigenvectors for a given $λ$ you solve ::: the null space $(A - λ I) x = 0$ Eigenvalues of a triangular matrix ::: its diagonal entries Sum of eigenvalues equals ::: the trace; product equals the determinant If $A x = λ x$ , then $A^{k} x =$ ::: $λ^{k} x$ $0$ is an eigenvalue iff ::: $A$ is singular ( $det A = 0$ )