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Lesson 114 — Eigenvalues and Eigenvectors

Invariant directions of a linear transformation: Av = λv. Characteristic polynomial, algebraic and geometric multiplicity. The cornerstone of PageRank, quantum mechanics, and PCA.

Used in: Linear Algebra (1st year engineering) · Equiv. Lineare Algebra LK German · Equiv. H2 Math Singapore · Math III advanced Japanese

A\vec{v} = \lambda\,\vec{v},\quad \vec{v} \neq \vec{0}

An eigenvector $\vec{v}$ of $A$ is a direction that $A$ only stretches or shrinks, without rotating. The scalar $\lambda$ is the eigenvalue. Eigenvalues are the roots of the characteristic polynomial $p_A(\lambda) = \det(A - \lambda I) = 0$ .

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Rigorous notation, full derivation, hypotheses

Rigorous definition

Eigenvalues and eigenvectors

Characteristic equation

Definition· Characteristic polynomial

The condition $A\vec{v} = \lambda\vec{v}$ with $\vec{v} \neq \vec{0}$ is equivalent to

$(A - \lambda I)\vec{v} = \vec{0}$

having a nontrivial solution, which occurs if and only if $A - \lambda I$ is singular:

p_A(\lambda) = \det(A - \lambda I) = 0.

what this means · Characteristic equation: the roots of p_A(λ) are exactly the eigenvalues of A.

$p_A$ is the characteristic polynomial of $A$ , of degree $n$ in $\lambda$ . Over $\mathbb{C}$ , it has exactly $n$ roots (with multiplicity).

"The characteristic polynomial of $A$ is $p_A(x) = \det(A - xI_n)$ . The eigenvalues of $A$ are exactly the roots of $p_A$ ." — Beezer, A First Course in Linear Algebra, §EE, Theorem EMRCP

Eigenspace and multiplicities

Fundamental properties

General vector rotates under A (yellow arrow deviates). Eigenvector only changes magnitude, remains on the same line (blue arrow).

Worked examples

Example— 1· Eigenvalues of 2x2 matrix via characteristic polynomial

Problem. Compute the eigenvalues and eigenvectors of $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$ .

Strategy. $A$ is upper triangular — eigenvalues are on the diagonal. Still we compute via polynomial to practice the general method.

Solution.

p_A(\lambda) = \det(A - \lambda I) = \det\begin{pmatrix} 3-\lambda & 1 \\ 0 & 2-\lambda \end{pmatrix} = (3-\lambda)(2-\lambda) = \lambda^2 - 5\lambda + 6.

Roots: $\lambda_1 = 3$ , $\lambda_2 = 2$ .

For $\lambda_1 = 3$ : $(A - 3I)\vec{v} = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}\vec{v} = \vec{0}$ $\Rightarrow$ $v_2 = 0$ , $v_1$ free. Eigenvector: $(1, 0)$ .

For $\lambda_2 = 2$ : $(A - 2I)\vec{v} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\vec{v} = \vec{0}$ $\Rightarrow$ $v_1 = -v_2$ . Eigenvector: $(-1, 1)$ .

Verification. $A(1,0) = (3,0) = 3(1,0)$ . $A(-1,1) = (-3+1, 2) = (-2, 2) = 2(-1, 1)$ . Correct.

Source. Beezer, A First Course in Linear Algebra, §EE, Example ESMS2 — GNU FDL.

Example— 2· Eigenvalues of 2x2 symmetric matrix (eigenvalues are real)

Problem. Compute the eigenvalues of $A = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$ and verify the properties $\operatorname{tr}(A) = \sum\lambda_i$ and $\det A = \prod\lambda_i$ .

Strategy. Degree-2 polynomial with quadratic formula. Then verification.

Solution.

$p_A(\lambda) = (5-\lambda)^2 - 16 = \lambda^2 - 10\lambda + 9 = (\lambda - 9)(\lambda - 1).$

Eigenvalues: $\lambda_1 = 9$ , $\lambda_2 = 1$ .

Verification: $\operatorname{tr}(A) = 5 + 5 = 10 = 9 + 1$ . $\det A = 25 - 16 = 9 = 9 \cdot 1$ . Both check out.

For $\lambda = 9$ : $(A - 9I)\vec{v} = \begin{pmatrix} -4 & 4 \\ 4 & -4 \end{pmatrix}\vec{v} = \vec{0}$ $\Rightarrow$ $v_1 = v_2$ . Eigenvector: $(1, 1)$ .

For $\lambda = 1$ : eigenvector $(1, -1)$ .

Verification. The two eigenvectors are orthogonal: $(1,1) \cdot (1,-1) = 0$ . Expected — $A$ is symmetric.

Source. Beezer, A First Course in Linear Algebra, §EE, Example ESMS4 — GNU FDL.

Example— 3· Algebraic vs geometric multiplicity (non-diagonalizable)

Problem. For $A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ , compute the characteristic polynomial, identify multiplicities, and decide if $A$ is diagonalizable.

Strategy. Compute $p_A$ , find roots, and dimension of eigenspace.

Solution.

$p_A(\lambda) = (2-\lambda)^2 \Rightarrow \lambda = 2$ with $m_a(2) = 2$ .

Eigenspace: $(A - 2I)\vec{v} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\vec{v} = \vec{0}$ $\Rightarrow$ $v_2 = 0$ , $v_1$ free. So $E_2 = \operatorname{span}\{(1,0)\}$ , with $m_g(2) = 1$ .

Since $m_g(2) = 1 < m_a(2) = 2$ , the matrix is not diagonalizable. Its Jordan form is $J = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ (the matrix itself, in this case).

Verification. If it were diagonalizable, there would exist two linearly independent eigenvectors — but there is only one. So there is no basis of eigenvectors for $\mathbb{R}^2$ .

Source. Beezer, A First Course in Linear Algebra, §EE, Example EMMS — GNU FDL.

Example— 4· Power of matrix via eigenvalues

Problem. Compute $A^{10}$ for $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ (Fibonacci matrix).

Strategy. Diagonalize $A = PDP^{-1}$ , then $A^{10} = PD^{10}P^{-1}$ .

Solution.

$p_A(\lambda) = \lambda^2 - \lambda - 1 = 0 \Rightarrow \lambda = \frac{1 \pm \sqrt{5}}{2}.$

Let $\phi = (1 + \sqrt{5})/2 \approx 1.618$ (golden ratio) and $\psi = (1 - \sqrt{5})/2 \approx -0.618$ .

Eigenvectors: for $\phi$ we have $(\phi, 1)$ ; for $\psi$ we have $(\psi, 1)$ .

$A^{10} = PD^{10}P^{-1}$ where $D = \operatorname{diag}(\phi^{10}, \psi^{10})$ .

In particular, the $(1,1)$ entry of $A^n$ is $F_{n+1}$ (the $(n+1)$ -th Fibonacci number), given by Binet's formula:

$F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}.$

Numerically: $\phi^{10} \approx 122.99$ , $\psi^{10} \approx 0.009$ , $F_{11} = 89$ .

Verification. $A^{10} = \begin{pmatrix} F_{11} & F_{10} \\ F_{10} & F_9 \end{pmatrix} = \begin{pmatrix} 89 & 55 \\ 55 & 34 \end{pmatrix}$ — confirmed by the Fibonacci sequence.

Source. Austin, Understanding Linear Algebra, §4.1 — CC-BY-SA.

Example— 5· Stationary distribution of Markov chain via eigenvector

Problem. A Markov chain models the weather in São Paulo: states Sunny (S) and Rainy (R). The transition matrix is

P = \begin{pmatrix} 0.8 & 0.2 \\ 0.4 & 0.6 \end{pmatrix}

where $P_{ij}$ is the probability of transitioning from state $j$ to state $i$ . Find the stationary distribution $\pi$ (eigenvector of $\lambda = 1$ ).

Strategy. Stochastic matrices always have $\lambda = 1$ as an eigenvalue. Solve $(P - I)\pi = \vec{0}$ with normalization $\pi_1 + \pi_2 = 1$ .

Solution.

$(P - I)\pi = \begin{pmatrix} -0.2 & 0.2 \\ 0.4 & -0.4 \end{pmatrix}\pi = \vec{0}$ $\Rightarrow$ $-0.2\pi_S + 0.2\pi_R = 0$ $\Rightarrow$ $\pi_S = \pi_R$ .

Normalizing: $\pi_S + \pi_R = 1 \Rightarrow \pi_S = \pi_R = 0.5$ .

In the long run, São Paulo is sunny 50% of the time and rainy 50%.

Verification. $P\pi = \begin{pmatrix} 0.8 \cdot 0.5 + 0.2 \cdot 0.5 \\ 0.4 \cdot 0.5 + 0.6 \cdot 0.5 \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix} = \pi$ . Correct.

Source. Austin, Understanding Linear Algebra, §4.3 — CC-BY-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 19Understanding 7Modeling 7Challenge 4Proof 3

Ex. 114.1Application
Compute the eigenvalues and eigenvectors of $A = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$ .
Select the correct option
A) $\lambda_1 = 3,\; \lambda_2 = 2$B) $\lambda_1 = 5,\; \lambda_2 = 0$C) $\lambda_1 = 6,\; \lambda_2 = 0$D) $\lambda_1 = 3,\; \lambda_2 = -2$
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Solve online Beezer, A First Course in Linear Algebra · §EE
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The matrix is diagonal, so eigenvalues are on the diagonal: $\lambda_1 = 3$ and $\lambda_2 = 2$ . Eigenvectors: $(1,0)$ and $(0,1)$ .
Show step-by-step (with the why)
1. For a diagonal matrix $\operatorname{diag}(a, b)$ , the characteristic polynomial is $(a-\lambda)(b-\lambda)$ . Why: the determinant of a diagonal matrix is the product of the diagonal.
2. Roots: $\lambda_1 = 3$ and $\lambda_2 = 2$ . No need for quadratic formula.
3. Eigenvectors: each coordinate axis. $A(1,0) = (3,0) = 3(1,0)$ .
Shortcut: diagonal matrix — eigenvalues are on the diagonal and eigenvectors are the canonical basis vectors.
Ex. 114.2Application
Compute the eigenvalues of $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ and find the corresponding eigenvectors.
Select the correct option
A) $\lambda_1 = 5,\; \lambda_2 = 2$B) $\lambda_1 = 7,\; \lambda_2 = 0$C) $\lambda_1 = 4,\; \lambda_2 = 3$D) $\lambda_1 = 6,\; \lambda_2 = 1$
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Polynomial: $\det(A - \lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda-5)(\lambda-2)$ . Eigenvalues: $\lambda_1 = 5$ , $\lambda_2 = 2$ . For $\lambda=5$ : eigenvector $(1,1)$ . For $\lambda=2$ : eigenvector $(1,-2)$ .
Ex. 114.3ApplicationAnswer key
Compute the eigenvalues of $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ .
Select the correct option
A) $\lambda_1 = 3,\; \lambda_2 = -1$B) $\lambda_1 = 3,\; \lambda_2 = 1$C) $\lambda_1 = 2,\; \lambda_2 = 0$D) $\lambda_1 = -3,\; \lambda_2 = 1$
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Polynomial: $(1-\lambda)^2 - 4 = \lambda^2 - 2\lambda - 3 = (\lambda-3)(\lambda+1)$ . Eigenvalues: $\lambda_1 = 3$ , $\lambda_2 = -1$ . Eigenvectors: $(1,1)$ and $(1,-1)$ .
Ex. 114.4Application
Compute the eigenvalues and eigenvectors of $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ .
Select the correct option
A) $\lambda_1 = 1,\; \lambda_2 = -1$B) $\lambda_1 = 0,\; \lambda_2 = 0$C) $\lambda_1 = 1,\; \lambda_2 = 1$D) $\lambda_1 = i,\; \lambda_2 = -i$
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Beezer, A First Course in Linear Algebra · §EE
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Polynomial: $(0-\lambda)^2 - 1 = \lambda^2 - 1 = (\lambda-1)(\lambda+1)$ . Eigenvalues: $\lambda_1 = 1$ and $\lambda_2 = -1$ . For $\lambda=1$ : $(1,1)$ . For $\lambda=-1$ : $(1,-1)$ . This is the reflection matrix in the angle bisector.
Ex. 114.5Application
Compute the eigenvalues of $A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$ .
Select the correct option
A) $\lambda_1 = 3,\; \lambda_2 = 2$B) $\lambda_1 = 4,\; \lambda_2 = 1$C) $\lambda_1 = 5,\; \lambda_2 = 0$D) $\lambda_1 = 6,\; \lambda_2 = -1$
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Polynomial: $(4-\lambda)(1-\lambda) + 2 = \lambda^2 - 5\lambda + 6 = (\lambda-3)(\lambda-2)$ . Eigenvalues: $3$ and $2$ . Verification: $\operatorname{tr} = 5 = 3+2$ ; $\det = 6 = 3 \cdot 2$ .
Show step-by-step (with the why)
1. Set up $A - \lambda I = \begin{pmatrix} 4-\lambda & -2 \\ 1 & 1-\lambda \end{pmatrix}$ .
2. Compute the determinant: $(4-\lambda)(1-\lambda) - (-2)(1) = \lambda^2 - 5\lambda + 6$ .
3. Factor: $(\lambda-3)(\lambda-2) = 0$ , roots $3$ and $2$ .
4. Check with sum and product of diagonal elements.
Mental shortcut: for $2 \times 2$ , use the formula $\lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = 0$ directly.
Ex. 114.6ApplicationAnswer key
Compute the eigenvalues and eigenvectors of $A = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$ .
Select the correct option
A) $\lambda_1 = 9,\; \lambda_2 = 1$B) $\lambda_1 = 5,\; \lambda_2 = 5$C) $\lambda_1 = 10,\; \lambda_2 = 0$D) $\lambda_1 = 9,\; \lambda_2 = -1$
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Polynomial: $(5-\lambda)^2 - 16 = \lambda^2 - 10\lambda + 9 = (\lambda-9)(\lambda-1)$ . Eigenvalues: $9$ and $1$ . Eigenvectors: $(1,1)$ and $(1,-1)$ , orthogonal (expected for symmetric).
Ex. 114.7ApplicationAnswer key
Analyze the diagonalizability of $A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ . Compute algebraic and geometric multiplicity.
Select the correct option
A) $m_a(2)=2,\; m_g(2)=1$ — not diagonalizableB) $m_a(2)=2,\; m_g(2)=2$ — diagonalizableC) $m_a(2)=1,\; m_g(2)=1$ — distinct eigenvaluesD) $m_a(2)=2,\; m_g(2)=0$ — no eigenvectors
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Upper triangular: eigenvalues on diagonal, $\lambda_1 = 2$ , $\lambda_2 = 2$ (multiplicity 2). Eigenspace: $(A - 2I) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ , $\dim = 1$ . So $m_g = 1 < m_a = 2$ — not diagonalizable.
Ex. 114.8ApplicationAnswer key
Compute the eigenvalues of $A = \begin{pmatrix} 6 & -1 \\ 2 & 3 \end{pmatrix}$ .
Select the correct option
A) $\lambda = \dfrac{9 \pm \sqrt{17}}{2}$B) $\lambda_1 = 5,\; \lambda_2 = 4$C) $\lambda = \dfrac{9 \pm \sqrt{97}}{2}$D) $\lambda = \dfrac{9 \pm \sqrt{17}}{4}$
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Solve online Austin, Understanding Linear Algebra · §4.1
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Polynomial: $\det(A-\lambda I) = (6-\lambda)(3-\lambda) - (-1)(2) = (6-\lambda)(3-\lambda) + 2 = \lambda^2 - 9\lambda + 20$ . Factoring: $(\lambda-5)(\lambda-4) = 0$ . Eigenvalues: $\lambda_1 = 5$ and $\lambda_2 = 4$ . Verification: $\operatorname{tr} = 9 = 5+4$ ; $\det = 20 = 5 \cdot 4$ .
Ex. 114.9Application
Compute the eigenvalues of $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ .
Select the correct option
A) $\lambda_1=1,\; \lambda_2=2,\; \lambda_3=3$B) $\lambda_1=1,\; \lambda_2=2,\; \lambda_3=0$C) $\lambda = 6$ (triple)D) $\lambda_1=0,\; \lambda_2=2,\; \lambda_3=3$
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Triangular: eigenvalues $1, 2, 3$ (on diagonal). Eigenvectors: $e_1, e_2, e_3$ of canonical basis.
Ex. 114.10Application
Compute the eigenvalues of $A = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 3 & 0 \\ 1 & 1 & 4 \end{pmatrix}$ .
Select the correct option
A) $\lambda_1=2,\; \lambda_2=3,\; \lambda_3=4$B) $\lambda_1=2,\; \lambda_2=4,\; \lambda_3=3$C) $\lambda = 9$ (triple)D) Real eigenvalues do not exist
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Lower triangular: eigenvalues on diagonal: $2, 3, 4$ . Verification: $\operatorname{tr} = 9 = 2+3+4$ ; $\det = 24 = 2\cdot3\cdot4$ .
Ex. 114.11ApplicationAnswer key
Compute the eigenvalues of $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and determine if it is diagonalizable.
Select the correct option
A) $\lambda=0$ with $m_a=2,\; m_g=1$ — not diagonalizableB) $\lambda=0$ with $m_a=2,\; m_g=2$ — diagonalizableC) $\lambda=0$ and $\lambda=1$ — two distinct eigenvaluesD) Real eigenvalues do not exist
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Nilpotent of order 2: $A^2 = 0$ . Polynomial: $(-\lambda)^2 = \lambda^2$ . Unique eigenvalue: $\lambda = 0$ with $m_a = 2$ . Eigenspace: $\ker A = \operatorname{span}\{(1,0)\}$ , $m_g = 1$ — not diagonalizable.
Ex. 114.12ApplicationAnswer key
Compute the eigenvalues of $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ .
Select the correct option
A) $\lambda_1=3,\; \lambda_2=\lambda_3=0$ — diagonalizableB) $\lambda_1=\lambda_2=\lambda_3=1$ — all equal to 1C) $\lambda_1=3,\; \lambda_2=\lambda_3=0$ — not diagonalizableD) $\lambda_1=3,\; \lambda_2=3,\; \lambda_3=0$
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Solve online Austin, Understanding Linear Algebra · §4.1
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Matrix of ones: $\operatorname{tr} = 3$ , $\det = 0$ . So $\lambda_1 = 3$ and $\lambda_2 = \lambda_3 = 0$ . Eigenvector of $\lambda=3$ : $(1,1,1)$ . Eigenspace of $\lambda=0$ : $\ker A$ , dimension 2. Diagonalizable.
Show step-by-step (with the why)
1. Note that every row of the matrix is $(1,1,1)$ , so $\operatorname{rank} = 1$ .
2. This implies $\dim\ker A = 2$ , so $\lambda = 0$ with $m_g \geq 2$ .
3. Sum of all eigenvalues = $\operatorname{tr} = 3$ , so the third eigenvalue is $3$ .
4. Check: $A(1,1,1)^T = (3,3,3)^T = 3(1,1,1)^T$ .
Curious fact: rank-1 matrices always have $n-1$ zero eigenvalues and one eigenvalue equal to the trace.
Ex. 114.13Application
If $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ has eigenvalues $1$ and $-1$ , what are the eigenvalues of $A^{10}$ ? Compute $A^{10}$ .
Select the correct option
A) Eigenvalues of $A^{10}$ are both $1$; thus $A^{10}=I$B) Eigenvalues of $A^{10}$ are $10$ and $-10$C) Eigenvalues of $A^{10}$ are $1$ and $-1$D) Eigenvalues of $A^{10}$ are $1$ and $0$
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If $A\vec{v} = \lambda\vec{v}$ , then $A^k\vec{v} = \lambda^k\vec{v}$ . Eigenvalues of $A^{10}$ : $1^{10} = 1$ and $(-1)^{10} = 1$ . So $A^{10}$ has eigenvalue $1$ with multiplicity 2, meaning $A^{10} = I$ .
Ex. 114.14Application
A matrix $A$ has eigenvalues $2$ and $3$ . What are the eigenvalues of $A^2 + I$ ?
Select the correct option
A) $5$ and $10$B) $4$ and $9$C) $3$ and $4$D) $5$ and $9$
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If eigenvalues of $A$ are $2$ and $3$ , then eigenvalues of $A^2$ are $4$ and $9$ , and eigenvalues of $A^2 + I$ are $5$ and $10$ . (Polynomial in matrix shifts eigenvalues.)
Ex. 114.15Application
A $3 \times 3$ matrix has eigenvalues $1$ , $2$ , $4$ . Compute $\det A$ and $\operatorname{tr}(A)$ .
Select the correct option
A) $\det A = 8,\; \operatorname{tr}(A) = 7$B) $\det A = 7,\; \operatorname{tr}(A) = 8$C) $\det A = 8,\; \operatorname{tr}(A) = 8$D) $\det A = 6,\; \operatorname{tr}(A) = 7$
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If eigenvalues are $1, 2, 4$ : $\det A = 1 \cdot 2 \cdot 4 = 8$ and $\operatorname{tr}(A) = 1+2+4 = 7$ .
Ex. 114.16Application
A $2 \times 2$ matrix has $\operatorname{tr}(A) = 5$ and $\det A = 6$ . Compute the eigenvalues.
Select the correct option
A) $\lambda_1 = 3,\; \lambda_2 = 2$B) $\lambda_1 = 5,\; \lambda_2 = 6$C) $\lambda_1 = 6,\; \lambda_2 = 5$D) $\lambda_1 = 3,\; \lambda_2 = -2$
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Solve online Austin, Understanding Linear Algebra · §4.1
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Polynomial: $\lambda^2 - \operatorname{tr}\lambda + \det = \lambda^2 - 5\lambda + 6 = (\lambda-3)(\lambda-2)$ . Eigenvalues $3$ and $2$ .
Ex. 114.17Application
Prove that if $\lambda$ is an eigenvalue of invertible $A$ , then $1/\lambda$ is an eigenvalue of $A^{-1}$ .
Select the correct option
A) The eigenvalue of $A^{-1}$ corresponding to $\lambda$ is $1/\lambda$B) The eigenvalue of $A^{-1}$ corresponding to $\lambda$ is $-\lambda$C) The eigenvalue of $A^{-1}$ corresponding to $\lambda$ is $\lambda$D) $A^{-1}$ does not have the same eigenvectors as $A$
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Beezer, A First Course in Linear Algebra · §PEE
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If $\lambda$ is an eigenvalue of invertible $A$ , then $1/\lambda$ is an eigenvalue of $A^{-1}$ . Proof: $A\vec{v} = \lambda\vec{v} \Rightarrow \vec{v} = \lambda A^{-1}\vec{v} \Rightarrow A^{-1}\vec{v} = (1/\lambda)\vec{v}$ .
Show step-by-step (with the why)
1. Start with $A\vec{v} = \lambda\vec{v}$ with $\lambda \neq 0$ (guaranteed since $A$ invertible).
2. Apply $A^{-1}$ to both sides: $\vec{v} = \lambda A^{-1}\vec{v}$ .
3. Divide by $\lambda$ : $A^{-1}\vec{v} = (1/\lambda)\vec{v}$ .
Remark: the same eigenvector $\vec{v}$ works for both $A$ and $A^{-1}$ ; only the eigenvalue changes to its reciprocal.
Ex. 114.18Application
Compute the eigenvalues of the rotation matrix $R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ for $\theta \in (0, \pi)$ .
Select the correct option
A) No real eigenvalues; complex eigenvalues $e^{\pm i\theta}$B) $\lambda = \cos\theta$ (double real eigenvalue)C) $\lambda = \pm 1$D) $\lambda = \cos\theta \pm \sin\theta$
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Rotation by $\theta \neq 0, \pi$ : polynomial $\lambda^2 - 2\cos\theta\,\lambda + 1 = 0$ . Discriminant: $4\cos^2\theta - 4 = -4\sin^2\theta < 0$ . No real eigenvalues. Complex eigenvalues: $e^{\pm i\theta}$ .
Ex. 114.19Understanding
Explain why a matrix with $\det A = 0$ necessarily has $0$ as an eigenvalue.
Select the correct option
A) $\det A=0$ implies $A\vec{v}=\vec{0}$ for some $\vec{v}\neq\vec{0}$, meaning $\lambda=0$ is an eigenvalueB) $\det A=0$ implies all eigenvalues are zeroC) $\det A=0$ implies $A=0$ (zero matrix)D) $\det A=0$ implies $\operatorname{tr}(A)=0$, which forces $\lambda=0$
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$\det A = 0$ means $A$ is singular, so $Ax = 0$ has a nontrivial solution $\vec{v} \neq 0$ . This means exactly $A\vec{v} = 0 = 0 \cdot \vec{v}$ , that is $\lambda = 0$ is an eigenvalue.
Ex. 114.20Understanding
Show that $A$ and $A^T$ have the same characteristic polynomial (and thus the same eigenvalues).
Select the correct option
A) $A$ and $A^T$ have the same eigenvalues but generally distinct eigenvectorsB) $A$ and $A^T$ have the same eigenvalues and the same eigenvectorsC) $A$ and $A^T$ have the same eigenvectors but eigenvalues can differD) $A$ and $A^T$ have completely different spectra
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$\det(A^T - \lambda I) = \det((A - \lambda I)^T) = \det(A - \lambda I)$ since $\det M = \det M^T$ for any matrix. So $A$ and $A^T$ have the same characteristic polynomial and thus the same eigenvalues. But the eigenvectors generally differ.
Ex. 114.21Understanding
If $B = P^{-1}AP$ (similar matrices), what can you conclude about the eigenvalues and eigenvectors of $A$ and $B$ ?
Select the correct option
A) Similar → same eigenvalues, same eigenvectorsB) Similar → same eigenvalues, eigenvectors related by PC) Similar → same trace but determinant can differD) Similar → same characteristic polynomial but eigenvalues can differ
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If $B = P^{-1}AP$ , then $\det(B - \lambda I) = \det(P^{-1}AP - \lambda I) = \det(P^{-1}(A-\lambda I)P) = \det(A - \lambda I)$ . Same polynomial, same eigenvalues. If $A\vec{v} = \lambda\vec{v}$ , then $B(P^{-1}\vec{v}) = \lambda(P^{-1}\vec{v})$ , so the eigenvectors of $B$ are $P^{-1}\vec{v}$ . Answer: B.
Ex. 114.22Understanding
If $A^2 = I$ , what are the only possible eigenvalues of $A$ ?
Select the correct option
A) Always $0$ and $1$ B) Always $+1$ and $-1$ C) Any pair of realsD) Depend on the specific matrix
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If $A^2 = I$ and $A\vec{v} = \lambda\vec{v}$ , then $\lambda^2\vec{v} = A^2\vec{v} = I\vec{v} = \vec{v}$ , so $\lambda^2 = 1$ and $\lambda = \pm 1$ . Answer: B.
Ex. 114.23Understanding
What are the eigenvalues of an orthogonal projection $P$ (with $P^2 = P$ )?
Select the correct option
A) Eigenvalues of orthogonal projection are always $0$ B) Eigenvalues of orthogonal projection are $0$ and $1$ C) Eigenvalues of orthogonal projection are $1$ and $-1$ D) Orthogonal projection has no real eigenvalues
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Solve online Austin, Understanding Linear Algebra · §4.1
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If $P$ is orthogonal projection, $P^2 = P$ . Then $\lambda^2 = \lambda$ , so $\lambda(\lambda-1) = 0$ : eigenvalues are $0$ or $1$ . Vectors in the projected subspace have $\lambda=1$ ; vectors in the complement have $\lambda=0$ . Answer: B.
Ex. 114.24Understanding
Prove that eigenvectors of distinct eigenvalues are linearly independent (case of two eigenvectors).
Select the correct option
A) Eigenvectors of distinct eigenvalues are linearly independentB) Eigenvectors of distinct eigenvalues can be parallel if the matrix is symmetricC) Linear independence depends on the dimension of the space, not on the eigenvaluesD) Eigenvectors of distinct eigenvalues are orthogonal in any matrix
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Beezer, A First Course in Linear Algebra · §PEE
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Suppose $A\vec{v}_1 = \lambda_1\vec{v}_1$ and $A\vec{v}_2 = \lambda_2\vec{v}_2$ with $\lambda_1 \neq \lambda_2$ . If $\alpha_1\vec{v}_1 + \alpha_2\vec{v}_2 = 0$ , apply $A$ : $\alpha_1\lambda_1\vec{v}_1 + \alpha_2\lambda_2\vec{v}_2 = 0$ . Subtract $\lambda_2$ times the first equation: $\alpha_1(\lambda_1-\lambda_2)\vec{v}_1 = 0$ . Since $\lambda_1 \neq \lambda_2$ and $\vec{v}_1 \neq 0$ : $\alpha_1 = 0$ , so $\alpha_2 = 0$ .
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1. Suppose linear dependence: $\alpha_1\vec{v}_1 + \alpha_2\vec{v}_2 = \vec{0}$ with $(\alpha_1, \alpha_2) \neq (0,0)$ .
2. Apply $A$ : $\alpha_1\lambda_1\vec{v}_1 + \alpha_2\lambda_2\vec{v}_2 = \vec{0}$ .
3. Subtract $\lambda_2$ times the original equation: $\alpha_1(\lambda_1 - \lambda_2)\vec{v}_1 = \vec{0}$ .
4. Since $\lambda_1 \neq \lambda_2$ and $\vec{v}_1 \neq \vec{0}$ : $\alpha_1 = 0$ , so $\alpha_2 = 0$ . Contradiction.
Shortcut: this argument generalizes by induction to any number of eigenvectors of distinct eigenvalues.
Ex. 114.25Understanding
Show that real eigenvalues of an orthogonal matrix $Q$ (with $Q^T Q = I$ ) satisfy $|\lambda| = 1$ .
Select the correct option
A) $|\lambda|=1$; possible real eigenvalues: $\pm 1$B) $|\lambda|\leq 1$; real eigenvalues can be any value in $[-1,1]$C) $\lambda=1$ necessarily for every orthogonal matrixD) Orthogonal matrices have no real eigenvalues
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Solve online Austin, Understanding Linear Algebra · §4.2
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Orthogonal matrices preserve norms: $\|Q\vec{v}\| = \|\vec{v}\|$ . If $Q\vec{v} = \lambda\vec{v}$ , then $|\lambda|\|\vec{v}\| = \|\lambda\vec{v}\| = \|Q\vec{v}\| = \|\vec{v}\|$ , so $|\lambda| = 1$ . Real eigenvalues: $\pm 1$ .
Ex. 114.26Modeling
A Markov chain of two regions (Southeast and Northeast) has transition matrix $P = \begin{pmatrix} 0.7 & 0.3 \\ 0.4 & 0.6 \end{pmatrix}$ . Find the stationary distribution via the eigenvector of $\lambda = 1$ .
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A) $\pi = (0.5,\; 0.5)$ — 50% in each region in the long runB) $\pi = (0.7,\; 0.3)$ — concentrates on the first row of the matrixC) $\pi = (0.4,\; 0.6)$ — concentrates on the second columnD) There is no stationary distribution for this matrix
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Solve online Austin, Understanding Linear Algebra · §4.3
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Transition matrix: $P = \begin{pmatrix} 0.7 & 0.3 \\ 0.4 & 0.6 \end{pmatrix}$ . Stationary distribution: eigenvector of $\lambda = 1$ . Solve $(P - I)\pi = 0$ : $-0.3\pi_1 + 0.3\pi_2 = 0$ → $\pi_1 = \pi_2$ . Normalize: $\pi = (0.5, 0.5)$ . Long run: 50% rainy, 50% sunny.
Show step-by-step (with the why)
1. Every stochastic matrix has $\lambda = 1$ as an eigenvalue. Why: the sum of each column is 1, so $\mathbf{1}^T P = \mathbf{1}^T$ .
2. Set up $(P - I)\pi = \vec{0}$ and row-reduce.
3. Normalize by the condition $\pi_1 + \pi_2 = 1$ .
4. Verify: $P\pi = \pi$ .
Curious fact: this is the foundation of Google's PageRank — the stationary distribution of the Markov chain where pages are states and links are transitions.
Ex. 114.27Modeling
The Fibonacci sequence is generated by $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ . Compute the eigenvalues and explain the growth of the sequence.
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A) $\phi=(1+\sqrt{5})/2$ and $\psi=(1-\sqrt{5})/2$; $|\phi|>1$ causes exponential growthB) $\lambda_1=1$ and $\lambda_2=0$; linear growthC) $\lambda_1=2$ and $\lambda_2=-1$; growth by $\lambda^n=2^n$D) $\lambda_1=\lambda_2=1$; constant growth
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Solve online Austin, Understanding Linear Algebra · §4.3
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Fibonacci polynomial: $p_A(\lambda) = \lambda^2 - \lambda - 1 = 0$ . Roots: $\phi = (1+\sqrt{5})/2 \approx 1.618$ (golden ratio) and $\psi = (1-\sqrt{5})/2 \approx -0.618$ . Since $|\phi| > 1$ , the sequence grows without bound. Since $|\psi| < 1$ , the $\\psi$-term decays to zero.
Ex. 114.28ModelingAnswer key
For the control system $\dot{x} = Ax$ with $A = \begin{pmatrix} -2 & 1 \\ 0 & -3 \end{pmatrix}$ : verify stability by analyzing the eigenvalues.
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A) Stable: eigenvalues $-2$ and $-3$, both with negative real partB) Unstable: eigenvalues $-2$ and $-3$, because they are negativeC) Marginally stable: eigenvalues real but not all negativeD) Undetermined: must compute the determinant first
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Solve online Austin, Understanding Linear Algebra · §4.1
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For $A = \begin{pmatrix} -2 & 1 \\ 0 & -3 \end{pmatrix}$ , eigenvalues are $-2$ and $-3$ (diagonal). Both have negative real part ( $-2 < 0$ and $-3 < 0$ ). So the system $\dot{x} = Ax$ is asymptotically stable — all solutions converge to zero.
Ex. 114.29Modeling
A Hessian matrix at a critical point is $H = \begin{pmatrix} -2 & 0 \\ 0 & -5 \end{pmatrix}$ . Identify the eigenvalues and classify the critical point (maximum/minimum/saddle).
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A) Eigenvalues $-2$ and $-5$: both negative — local maximum pointB) Eigenvalues $-2$ and $-5$: both negative — local minimum pointC) Eigenvalues $-2$ and $-5$: saddle (opposite signs)D) Eigenvalues $2$ and $5$: positive, thus local minimum point
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Solve online Beezer, A First Course in Linear Algebra · §PEE
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Given $A = \begin{pmatrix} -2 & 0 \\ 0 & -5 \end{pmatrix}$ (diagonal, symmetric): eigenvalues $-2$ and $-5$ , both negative. So $A$ is negative definite. This means the system $\dot{x} = Ax$ is stable and $-A$ is positive definite (would be a Hessian of a minimum).
Ex. 114.30ModelingAnswer key
For a path graph with 3 nodes (1—2—3), set up the Laplacian $L = D - W$ , compute the eigenvalues, and identify the number of connected components.
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A) Eigenvalues $0,1,3$; multiplicity of $\lambda=0$ is 1, thus 1 connected componentB) Eigenvalues $0,0,3$; multiplicity of $\lambda=0$ is 2, thus 2 componentsC) Eigenvalues $1,2,3$; all positive, thus the graph is completeD) Eigenvalues $0,2,3$; multiplicity of $\lambda=0$ is 1, thus 1 component
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Solve online Austin, Understanding Linear Algebra · §4.3
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Laplacian of path graph with 3 nodes: $L = \begin{pmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{pmatrix}$ . Eigenvalues: $0, 1, 3$ . The eigenvalue $\lambda = 0$ has multiplicity 1 — confirms the graph has 1 connected component. Eigenvector of $\lambda=0$ : $(1,1,1)$ .
Show step-by-step (with the why)
1. Set up the Laplacian $L = D - W$ where $D$ is diagonal of degrees and $W$ the adjacency matrix.
2. For the path 1-2-3: nodes 1 and 3 have degree 1; node 2 has degree 2.
3. Compute $\det(L - \lambda I) = 0$ .
4. Observe: $L$ always has $\lambda = 0$ (sum of rows is zero). Number of components = multiplicity of $\lambda = 0$ .
Curious fact: the second smallest eigenvalue of the Laplacian (Fiedler value) measures how well-connected the graph is — foundation of spectral clustering in machine learning.
Ex. 114.31ModelingAnswer key
Prove that if $\lambda$ is an eigenvalue of $A$ with eigenvector $\vec{v}$ , then $\lambda + c$ is an eigenvalue of $A + cI$ with the same eigenvector $\vec{v}$ .
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A) The eigenvalue corresponding to $\lambda$ in $A+cI$ is $\lambda+c$, with the same eigenvectorB) The eigenvalue corresponding to $\lambda$ in $A+cI$ is $\lambda \cdot c$C) The eigenvalue corresponding to $\lambda$ in $A+cI$ is $\lambda$; only the eigenvector changesD) $A+cI$ has no eigenvalues related to those of $A$
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Austin, Understanding Linear Algebra · §4.2
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If $\lambda$ is an eigenvalue of $A$ with eigenvector $\vec{v}$ , then for $A + cI$ : $(A+cI)\vec{v} = A\vec{v} + c\vec{v} = \lambda\vec{v} + c\vec{v} = (\lambda+c)\vec{v}$ . So $\lambda + c$ is an eigenvalue of $A + cI$ , with the same eigenvector.
Ex. 114.32Modeling
In finance, the covariance matrix of two identical stocks with variance $\sigma^2$ and correlation $\rho$ is $\Sigma = \sigma^2 \begin{pmatrix} 1 & \rho \\ \rho & 1 \end{pmatrix}$ . Compute the eigenvalues and interpret.
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A) $\sigma^2(1+\rho)$ and $\sigma^2(1-\rho)$; common factor $(1,1)$, differential $(1,-1)$B) $\sigma^2$ and $\sigma^2\rho$; eigenvalues equal to diagonal and off-diagonalC) $1+\rho$ and $1-\rho$ (without the $\sigma^2$ factor)D) $\sigma^2(1+\rho)$ and $\sigma^2(1+\rho)$ — double eigenvalue
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Solve online Beezer, A First Course in Linear Algebra · §PEE
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The covariance matrix of two stocks with equal variance $\sigma^2$ and correlation $\rho$ is $\Sigma = \sigma^2\begin{pmatrix}1 & \rho \\ \rho & 1\end{pmatrix}$ . Eigenvalues: $\sigma^2(1+\rho)$ and $\sigma^2(1-\rho)$ . Eigenvectors: $(1,1)$ (common factor) and $(1,-1)$ (differential factor).
Ex. 114.33Challenge
Prove that if $\lambda$ is an eigenvalue of $A$ with eigenvector $\vec{v}$ , then $\lambda^k$ is an eigenvalue of $A^k$ for every positive integer $k$ .
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A) $\lambda^k$ is an eigenvalue of $A^k$ with the same eigenvector $\vec{v}$B) $k\lambda$ is an eigenvalue of $A^k$ with the same eigenvectorC) $\lambda^k$ is an eigenvalue of $A^k$ but the eigenvector changes at each stepD) $\lambda^k$ is an eigenvalue only if $A$ is symmetric
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Beezer, A First Course in Linear Algebra · §PEE
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Let $A\vec{v} = \lambda\vec{v}$ . Apply $A^2$ : $A^2\vec{v} = A(A\vec{v}) = A(\lambda\vec{v}) = \lambda A\vec{v} = \lambda^2\vec{v}$ . By induction: $A^k\vec{v} = \lambda^k\vec{v}$ . So $\lambda^k$ is an eigenvalue of $A^k$ with the same eigenvector $\vec{v}$ .
Ex. 114.34ChallengeAnswer key
Prove that eigenvalues of an idempotent matrix ( $A^2 = A$ ) are only $0$ or $1$ .
Select the correct option
A) Only $0$ and $1$ are possible, since $\lambda^2=\lambda$ implies $\lambda(\lambda-1)=0$B) Only $1$ is possible; $\lambda=0$ would imply the zero matrixC) Any $\lambda$ satisfies $A^2=A$ when $A$ has full rankD) Only $-1$ and $1$ are possible, by analogy with $A^2=I$
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Beezer, A First Course in Linear Algebra · §PEE
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If $A$ is idempotent ( $A^2 = A$ ) and $A\vec{v} = \lambda\vec{v}$ , then $\lambda^2\vec{v} = A^2\vec{v} = A\vec{v} = \lambda\vec{v}$ , so $\lambda^2 = \lambda$ and $\lambda \in \{0, 1\}$ . Examples: orthogonal projections are idempotent.
Show step-by-step (with the why)
1. Use $A^2\vec{v} = A(\lambda\vec{v}) = \lambda^2\vec{v}$ (result of previous exercise).
2. Use $A^2 = A$ : thus $A^2\vec{v} = A\vec{v} = \lambda\vec{v}$ .
3. Combine: $\lambda^2\vec{v} = \lambda\vec{v}$ , that is $(\lambda^2 - \lambda)\vec{v} = \vec{0}$ .
4. Since $\vec{v} \neq \vec{0}$ : $\lambda(\lambda-1) = 0$ .
Remark: idempotent matrices are exactly the projections (not necessarily orthogonal).
Ex. 114.35Challenge
Construct a $2 \times 2$ matrix with eigenvalues $1$ and $-1$ such that $(1, 1)$ is an eigenvector of $\lambda = 1$ and $(1, -1)$ is an eigenvector of $\lambda = -1$ .
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A) $A = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$B) $A = \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$C) $A = \begin{pmatrix}1 & 1\\ 1 & -1\end{pmatrix}$D) $A = \begin{pmatrix}0 & -1\\ -1 & 0\end{pmatrix}$
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Solve online Beezer, A First Course in Linear Algebra · §PEE
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Find $2 \times 2$ with eigenvalues $\lambda_1 = 1$ , $\lambda_2 = -1$ and eigenvectors $\vec{v}_1 = (1,1)$ , $\vec{v}_2 = (1,-1)$ . Construct $A = PDP^{-1}$ where $P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ , $D = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ . Result: $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ .
Ex. 114.36Challenge
Prove that eigenvectors of a symmetric matrix corresponding to distinct eigenvalues are orthogonal.
Select the correct option
A) For symmetric matrix: eigenvectors of distinct eigenvalues are orthogonalB) For any matrix: eigenvectors of distinct eigenvalues are orthogonalC) For symmetric matrix: eigenvectors are orthogonal even if eigenvalues are equalD) Orthogonality of eigenvectors holds only for rotation matrices
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Austin, Understanding Linear Algebra · §4.2
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If $A$ symmetric and $A\vec{v}_1 = \lambda_1\vec{v}_1$ , $A\vec{v}_2 = \lambda_2\vec{v}_2$ with $\lambda_1 \neq \lambda_2$ : $\lambda_1(\vec{v}_1 \cdot \vec{v}_2) = (A\vec{v}_1) \cdot \vec{v}_2 = \vec{v}_1 \cdot (A\vec{v}_2) = \lambda_2(\vec{v}_1 \cdot \vec{v}_2)$ . Thus $(\lambda_1 - \lambda_2)(\vec{v}_1 \cdot \vec{v}_2) = 0$ . Since $\lambda_1 \neq \lambda_2$ : $\vec{v}_1 \cdot \vec{v}_2 = 0$ .
Ex. 114.37Proof
Prove that a triangular matrix (upper or lower) has eigenvalues equal to the diagonal elements.
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A) $p_A(\lambda) = \prod_{i=1}^n (a_{ii}-\lambda)$; eigenvalues are the diagonal elementsB) $p_A(\lambda) = \sum_{i=1}^n (a_{ii}-\lambda)$; eigenvalues are the diagonal elementsC) $p_A(\lambda) = \prod_{i=1}^n (a_{ii}-\lambda)$; but eigenvalues are the sum of rowsD) Only holds for diagonal: strictly triangular matrices don't have this property
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Beezer, A First Course in Linear Algebra · §EE
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Upper triangular matrix $A$ with diagonal $a_{11}, \ldots, a_{nn}$ . The determinant of $A - \lambda I$ (also triangular) is the product of the diagonal: $p_A(\lambda) = (a_{11}-\lambda)(a_{22}-\lambda)\cdots(a_{nn}-\lambda)$ . The roots are exactly the diagonal elements.
Show step-by-step (with the why)
1. Note that $A - \lambda I$ is triangular (upper or lower) with diagonal $(a_{ii} - \lambda)$ . Why: subtracting $\lambda$ from the diagonal of a triangular matrix doesn't change triangular structure.
2. The determinant of a triangular matrix is the product of diagonal elements: $\det(A-\lambda I) = \prod_{i=1}^n (a_{ii} - \lambda)$ .
3. The roots of $\prod (a_{ii}-\lambda) = 0$ are $\lambda = a_{ii}$ for $i = 1,\ldots,n$ .
Shortcut: this result holds for upper and lower triangular. Applies immediately to diagonal matrices as a special case.
Ex. 114.38Proof
Prove (by induction) that eigenvectors corresponding to $k$ distinct eigenvalues are linearly independent.
Select the correct option
A) The induction proof applies $(A-\lambda_k I)$ to the zero combination to eliminate $\vec{v}_k$ and use the IHB) The proof uses the Gram determinant to show the eigenvector matrix is invertibleC) The proof follows directly from the fact that eigenvectors form an orthonormal basisD) The proof uses only the case $k=2$; the general case is trivial by symmetry
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Beezer, A First Course in Linear Algebra · §PEE
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Let $\lambda_1, \ldots, \lambda_k$ be eigenvalues of $A$ with corresponding eigenvectors $\vec{v}_1, \ldots, \vec{v}_k$ , all distinct. Prove by induction on $k$ . Base $k=1$ : one nonzero eigenvector is LI by definition. Step: assume $\vec{v}_1, \ldots, \vec{v}_{k-1}$ are LI. If $\sum_{i=1}^k \alpha_i \vec{v}_i = 0$ , apply $A - \lambda_k I$ and use the induction hypothesis to conclude $\alpha_i = 0$ for all $i$ .
Ex. 114.39Proof
Prove that every real symmetric matrix has only real eigenvalues.
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A) The argument shows $\bar{\lambda}=\lambda$ using symmetry of $A$ and complex inner productB) The argument uses that the characteristic polynomial of real $A$ has only real roots in generalC) The argument uses that $\det(A-\lambda I)$ has real coefficients, thus roots are realD) The argument uses $\operatorname{tr}(A)$ and $\det(A)$ to exclude complex roots
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Beezer, A First Course in Linear Algebra · §PEE
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For $A$ symmetric and eigenvalue $\lambda$ : suppose $\lambda \in \mathbb{C}$ , $A\vec{v} = \lambda\vec{v}$ . Take the conjugate: $A\bar{\vec{v}} = \bar{\lambda}\bar{\vec{v}}$ (since $A$ is real). Then $\bar{\lambda}(\vec{v} \cdot \bar{\vec{v}}) = (A\vec{v}) \cdot \bar{\vec{v}} = \vec{v} \cdot (A\bar{\vec{v}}) = \lambda(\vec{v} \cdot \bar{\vec{v}})$ . Since $\|\vec{v}\|^2 = \vec{v} \cdot \bar{\vec{v}} > 0$ : $\bar{\lambda} = \lambda$ , so $\lambda \in \mathbb{R}$ .
Show step-by-step (with the why)
1. Suppose $\lambda \in \mathbb{C}$ and $A\vec{v} = \lambda\vec{v}$ .
2. Take the inner product of both sides with $\bar{\vec{v}}$ : $\langle A\vec{v}, \bar{\vec{v}}\rangle = \lambda\|\vec{v}\|^2$ .
3. Use symmetry of $A$ : $\langle A\vec{v}, \bar{\vec{v}}\rangle = \langle\vec{v}, A\bar{\vec{v}}\rangle = \bar{\lambda}\|\vec{v}\|^2$ .
4. Therefore $\lambda = \bar{\lambda}$ , that is $\lambda \in \mathbb{R}$ .
Remark: this is the spectral theorem for real symmetric matrices — they always have real eigenvalues and orthogonal eigenvectors (see Lesson 116).
Ex. 114.40Application
For $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$ with characteristic polynomial $(4-\lambda)(1-\lambda)^2$ , determine all eigenvalues, their algebraic and geometric multiplicities, and decide if $A$ is diagonalizable.
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Eigenvalues: $\lambda = 4$ (simple) and $\lambda = 1$ (double); eigenspace of $\lambda=1$ has dimension 2Eigenvalues: $\lambda = 4$ (triple); eigenspace has dimension 1Eigenvalues: $\lambda = 1$ (simple) and $\lambda = 4$ (double); eigenspace of $\lambda=4$ has dimension 1Eigenvalues: $\lambda = 0$ , $\lambda = 1$ , $\lambda = 4$ ; all simple
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Solve online Beezer, A First Course in Linear Algebra · §EE
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Characteristic polynomial given: $p_A(\lambda) = (4-\lambda)(1-\lambda)^2$ . Eigenvalues: $\lambda_1 = 4$ (algebraic multiplicity 1) and $\lambda_2 = 1$ (algebraic multiplicity 2). For $\lambda = 1$ : $(A - I)$ has rank 1 (since $A = \begin{pmatrix}2&1&1\\1&2&1\\1&1&2\end{pmatrix}$ and $A - I = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$ ). So kernel dimension = 2, meaning $m_g(1) = 2 = m_a(1)$ . For $\lambda = 4$ : eigenvector $(1,1,1)^T$ .
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1. Polynomial given: $(4-\lambda)(1-\lambda)^2 = 0$ . Roots: $\lambda_1 = 4$ , $\lambda_2 = 1$ (double). Why: read the roots directly from the factored polynomial.
2. For $\lambda = 1$ : compute $A - I$ : each entry is $a_{ij} - \delta_{ij}$ . Result: matrix in which each row is $(1,1,1)$ . Rank = 1, so kernel has dimension $3 - 1 = 2$ .
3. Since $m_g(1) = 2 = m_a(1)$ and $m_g(4) = 1 = m_a(4)$ , we have 3 linearly independent eigenvectors — the matrix is diagonalizable.
Shortcut: to check the dimension of eigenspace, compute rank of $A - \lambda I$ by row reduction; $m_g = n - \text{rank}$ .

Sources

A First Course in Linear Algebra — Robert A. Beezer · 2022 · EN · GNU FDL · §EE and §PEE. Primary source of exercises and rigorous definitions.
Understanding Linear Algebra — David Austin · 2023 · EN · CC-BY-SA · §4.1–§4.3. Source of geometric examples and applications to Markov chains.
Linear Algebra Done Right (4th ed.) — Sheldon Axler · 2024 · EN · CC-BY-NC · Ch. 5. Reference for modern approach to multiplicities and eigenspaces.