MAT1041 · Week 13 · - Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors

You learn to set up the characteristic equation and find eigenvalues, to compute a basis of the eigenspace for each eigenvalue, to tell algebraic and geometric multiplicity apart, and the shortcuts relating eigenvalues to trace, determinant, powers and inverse.

eigenvalueeigenvectorcharacteristic polynomialcharacteristic equationeigenspacealgebraic multiplicitygeometric multiplicitytriangular matrixtracedeterminant

When a matrix multiplies a vector, it generally both rotates and stretches it. But there are some special vectors that the matrix does not rotate, it only multiplies by a number. These are the eigenvectors, and the multiplier is the eigenvalue. This week we learn to find them; on the next page, if there are enough eigenvectors, we will make the matrix diagonal. The computing side is completely familiar: determinants (weeks 4-5) and the null space (week 9).

Golden rules:

First the eigenvalue: $\det(\lambda I - A) = 0$ .
Then for each eigenvalue the eigenspace: solve $(A - \lambda I)\mathbf{x} = \mathbf{0}$ .
$A - \lambda I$ must come out singular. If there is no zero row, you have found the eigenvalue wrongly.
Check: the sum of the eigenvalues is the trace, their product is the determinant.

Definition

Let $A$ be $n \times n$ . If, for a nonzero $\mathbf{v}$ and a number $\lambda$ ,

$A\mathbf{v} = \lambda\mathbf{v}$

then $\lambda$ is an eigenvalue of $A$ and $\mathbf{v}$ is an eigenvector corresponding to $\lambda$ .

$\mathbf{v} = \mathbf{0}$ is never counted as an eigenvector ( $A\mathbf{0} = \lambda\mathbf{0}$ for every $\lambda$ , which would be meaningless).
$\lambda = 0$ can be an eigenvalue: it means $A\mathbf{v} = \mathbf{0}$ .
If $\mathbf{v}$ is an eigenvector, so is every $c\mathbf{v}$ ( $c \neq 0$ ). Eigenvectors are not single, they come along a line (or subspace).

Geometry. $A\mathbf{v}$ lies on the same line as $\mathbf{v}$ . If $\lambda > 1$ it stretches, if $0 < \lambda < 1$ it shrinks, if $\lambda < 0$ it reverses direction. For the projection onto the $x$ -axis in the plane, $\mathbf{e}_1$ corresponds to eigenvalue 1 and $\mathbf{e}_2$ to eigenvalue 0.

The characteristic equation

$A\mathbf{v} = \lambda\mathbf{v} \iff (\lambda I - A)\mathbf{v} = \mathbf{0}$ . For a nonzero solution to exist, $\lambda I - A$ must be singular:

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

The left side is a polynomial of degree $n$ in $\lambda$ : the characteristic polynomial. Its roots are exactly the eigenvalues. (Using $\det(A - \lambda I)$ gives the same roots; the sign differs only if $n$ is odd.)

Example 1. $A = \begin{bmatrix} 2 & 3 \\ 2 & 1 \end{bmatrix}$ .

$\det(\lambda I - A) = \begin{vmatrix} \lambda - 2 & -3 \\ -2 & \lambda - 1 \end{vmatrix} = (\lambda - 2)(\lambda - 1) - 6 = \lambda^2 - 3\lambda - 4 = (\lambda - 4)(\lambda + 1).$

The eigenvalues are $4$ and $-1$ . The $2 \times 2$ shortcut: $\lambda^2 - (\operatorname{tr}A)\lambda + \det A$ , here trace 3, det $2 - 6 = -4$ .

The eigenspace

All the eigenvectors for a given $\lambda$ plus $\mathbf{0}$ form the null space of $A - \lambda I$ . It is called the eigenspace $E_\lambda$ of $\lambda$ ; it is a subspace.

Example 1 continued:

$\lambda = 4: \quad A - 4I = \begin{bmatrix} -2 & 3 \\ 2 & -3 \end{bmatrix} \xrightarrow{R_2 \leftarrow R_2 + R_1} \begin{bmatrix} -2 & 3 \\ 0 & 0 \end{bmatrix} \Rightarrow -2x + 3y = 0, \quad E_4 = \operatorname{span}\{(3, 2)\}.$

$\lambda = -1: \quad A + I = \begin{bmatrix} 3 & 3 \\ 2 & 2 \end{bmatrix} \xrightarrow{R_2 \leftarrow R_2 - \frac23 R_1} \begin{bmatrix} 3 & 3 \\ 0 & 0 \end{bmatrix} \Rightarrow x + y = 0, \quad E_{-1} = \operatorname{span}\{(1, -1)\}.$

Check: $A(3, 2)^T = (6 + 6, 6 + 2) = (12, 8) = 4(3, 2)$ . $A(1, -1)^T = (2 - 3, 2 - 1) = (-1, 1) = -1 \cdot (1, -1)$ .

PITFALL · getting a unique solution. If solving $(A - \lambda I)\mathbf{x} = \mathbf{0}$ gives only $\mathbf{x} = \mathbf{0}$ , there is a calculation error in the eigenvalue. At a true eigenvalue $A - \lambda I$ is always singular and at least one zero row appears. This is both a warning and a free check.

A 3×3 example and multiplicity

Example 2. $A = \begin{bmatrix} 3 & 2 & -2 \\ 2 & 3 & -2 \\ 2 & 2 & -1 \end{bmatrix}$ .

Characteristic polynomial (the calculation is long; using the common structure in the first two columns or expanding along the first row helps): $\lambda^3 - 5\lambda^2 + 7\lambda - 3 = (\lambda - 1)^2(\lambda - 3)$ . Check: trace $3 + 3 - 1 = 5 = 1 + 1 + 3$ , det $= 1 \cdot 1 \cdot 3 = 3$ .

$\lambda = 1: \quad A - I = \begin{bmatrix} 2 & 2 & -2 \\ 2 & 2 & -2 \\ 2 & 2 & -2 \end{bmatrix} \to \begin{bmatrix} 1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}: \quad x = -y + z,$

$E_1 = \operatorname{span}\{(-1, 1, 0),\ (1, 0, 1)\} \quad (\text{dimension } 2).$

$\lambda = 3: \quad A - 3I = \begin{bmatrix} 0 & 2 & -2 \\ 2 & 0 & -2 \\ 2 & 2 & -4 \end{bmatrix} \xrightarrow[R_1 \leftarrow \frac12R_1,\ R_2 \leftarrow R_2 - 2R_1]{R_1 \leftrightarrow R_3} \begin{bmatrix} 1 & 1 & -2 \\ 0 & -2 & 2 \\ 0 & 2 & -2 \end{bmatrix} \xrightarrow{R_3 \leftarrow R_3 + R_2} \begin{bmatrix} 1 & 1 & -2 \\ 0 & -2 & 2 \\ 0 & 0 & 0 \end{bmatrix}$

$y = z$ , $x = -y + 2z = z$ : $E_3 = \operatorname{span}\{(1, 1, 1)\}$ .

The two multiplicities:

	Definition	$\lambda = 1$ in Example 2	$\lambda = 2$ in Q7
Algebraic	the multiplicity of the root in the characteristic polynomial	2	2
Geometric	$\dim E_\lambda$	2	1

Always $1 \le \text{geometric} \le \text{algebraic}$ . Whether the two are equal is the key to diagonalization (next page).

Shortcuts

If the eigenvalues of $A$ are $\lambda_1, \dots, \lambda_n$ (with multiplicities):

Information	Result
$\lambda_1 + \dots + \lambda_n$	$\operatorname{tr}A$
$\lambda_1\lambda_2\cdots\lambda_n$	$\det A$
$A$ triangular	eigenvalues on the diagonal
$0$ is an eigenvalue	$A$ is singular
$A^k$	eigenvalues $\lambda^k$ , same eigenvectors
$A^{-1}$	$1/\lambda$ , same eigenvectors
$A + cI$	$\lambda + c$
$A^T$	same eigenvalues (eigenvectors may differ)

One more item for the invertible matrix theorem: $A$ is invertible $\iff$ 0 is not an eigenvalue of $A$ .

There may be no real eigenvalue

The characteristic polynomial of the $90^\circ$ rotation matrix $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ is $\lambda^2 + 1$ : no real root. In this course we usually work with real eigenvalues; in week 14 we will see that the eigenvalues of symmetric matrices are always real.

Common mistakes

Wrong	Right
Writing $\det(A - \lambda)$	$\det(A - \lambda I)$ : $\lambda$ only on the diagonal
Counting $\mathbf{v} = \mathbf{0}$ as an eigenvector	An eigenvector must be nonzero
Expanding the characteristic polynomial and losing the factors	Keep the common factor, take $(\lambda - 3)$ out
Thinking the eigenspace is the null space of $A$	It is the null space of $A - \lambda I$
Writing a single vector for the eigenspace basis and forgetting the second free variable	As many vectors as free variables
Thinking row operations preserve eigenvalues	Row operations change eigenvalues; don’t reduce $A$ first
Assuming algebraic and geometric multiplicity are equal	Compute; they differ in Q7

Test yourself

Exam-style questions. Solve them on paper first, then reveal the answer.

Question 1easy

Show that $\mathbf{v} = (1, 1)$ is an eigenvector of $A = \begin{bmatrix} 3 & 1 \\ 2 & 2 \end{bmatrix}$ and find the corresponding eigenvalue.

Show answer

\lambda = 4

$A\mathbf{v} = (3 + 1,\ 2 + 2) = (4, 4) = 4(1, 1)$ . So $A\mathbf{v} = 4\mathbf{v}$ , $\mathbf{v} \neq \mathbf{0}$ : an eigenvector, with eigenvalue 4.

Question 2easy

Find the eigenvalues of $\begin{bmatrix} 2 & 5 & -1 \\ 0 & -3 & 4 \\ 0 & 0 & 7 \end{bmatrix}$ .

Show answer

2,\ -3,\ 7

A triangular matrix: $\lambda I - A$ is triangular too, and its determinant is the product of the diagonal $(\lambda - 2)(\lambda + 3)(\lambda - 7)$ .
The eigenvalues of a triangular matrix are its diagonal entries.

Question 3medium

Find the characteristic polynomial and the eigenvalues of $A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}$ .

Show answer

\lambda^2 - 4\lambda - 5

;

\lambda = 5, -1

$\det(\lambda I - A) = (\lambda - 1)(\lambda - 3) - 8 = \lambda^2 - 4\lambda - 5 = (\lambda - 5)(\lambda + 1)$ .
Shortcut ( $2 \times 2$ ): $\lambda^2 - (\operatorname{tr}A)\lambda + \det A$ ; trace 4, det $3 - 8 = -5$ .

Question 4medium

For $A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}$ , find a basis for each eigenspace.

Show answer

E_5

\{(1, 1)\}

;

E_{-1}

\{(-2, 1)\}

$\lambda = 5$ : $A - 5I = \begin{bmatrix} -4 & 4 \\ 2 & -2 \end{bmatrix}$ , the rows are proportional, $x = y$ : $(1, 1)$ . $\lambda = -1$ : $A + I = \begin{bmatrix} 2 & 4 \\ 2 & 4 \end{bmatrix}$ , $x + 2y = 0$ : $(-2, 1)$ .
Check: $A(-2, 1)^T = (-2 + 4, -4 + 3) = (2, -1) = -1 \cdot (-2, 1)$ .

Question 5medium

Find the eigenvalues of $\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$ . What does one of them tell you about the matrix?

Show answer

0

and

5

; the matrix is singular

Trace 5, det $4 - 4 = 0$ : $\lambda^2 - 5\lambda = \lambda(\lambda - 5)$ . $\lambda = 0$ is an eigenvalue $\iff$ $A\mathbf{v} = \mathbf{0}$ has a nonzero solution $\iff$ $A$ is singular.

Question 6medium

Find the eigenvalues of $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 3 & 0 \\ 2 & 0 & 1 \end{bmatrix}$ with their algebraic multiplicities.

Show answer

\lambda = 3

(multiplicity 2),

\lambda = -1

(multiplicity 1)

Expand $\det(\lambda I - A)$ along row 2 (its only entry is $\lambda - 3$ ): $(\lambda - 3)\left[(\lambda - 1)^2 - 4\right]$ . $(\lambda - 1)^2 - 4 = (\lambda - 3)(\lambda + 1)$ . Total: $(\lambda - 3)^2(\lambda + 1)$ .
Check: the sum of the eigenvalues $3 + 3 - 1 = 5 = \operatorname{tr}A$ .

Question 7hard

For $A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ , find the algebraic and geometric multiplicity of $\lambda = 2$ .

Show answer

Algebraic 2, geometric 1

Triangular: the eigenvalues are $2, 2, 3$ . Algebraic multiplicity 2. $A - 2I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ : $y = 0$ , $z = 0$ , $x$ free. $E_2 = \operatorname{span}\{(1, 0, 0)\}$ , dimension 1.
The geometric multiplicity is always $\le$ the algebraic; here it is smaller.

Question 8medium

$\lambda = 3$ is an eigenvalue of an invertible matrix $A$ with eigenvector $\mathbf{v}$ . Find the corresponding eigenvalues of $A^2 - 2A + I$ and of $A^{-1}$ .

Show answer

4

and

1/3

$A^2\mathbf{v} = 9\mathbf{v}$ , so $(A^2 - 2A + I)\mathbf{v} = (9 - 6 + 1)\mathbf{v} = 4\mathbf{v}$ .
Multiply $A\mathbf{v} = 3\mathbf{v}$ from the left by $A^{-1}$ : $\mathbf{v} = 3A^{-1}\mathbf{v}$ , $A^{-1}\mathbf{v} = \frac13\mathbf{v}$ .
The eigenvector stays the same.

Question 9medium

A $3 \times 3$ matrix $A$ has eigenvalues $1, 2, -3$ . Find $\det A$ and $\operatorname{tr}A$ .

Show answer

\det A = -6

\operatorname{tr}A = 0

Determinant = the product of the eigenvalues: $1 \cdot 2 \cdot (-3) = -6$ .
Trace = the sum of the eigenvalues: $1 + 2 - 3 = 0$ .

Question 10hard

Does $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ (rotation by $90^\circ$ ) have real eigenvalues?

Show answer

No (the eigenvalues are

\pm i

)

$\det(\lambda I - A) = \lambda^2 + 1$ . It has no real root.
Geometric meaning: a $90^\circ$ rotation leaves no nonzero vector along its own direction.

Question 11hard

Find $k$ such that $\lambda = 4$ is an eigenvalue of $\begin{bmatrix} 2 & k \\ 1 & 3 \end{bmatrix}$ .

Show answer

k = 2

$\lambda = 4$ is an eigenvalue $\iff \det(A - 4I) = 0$ . $\begin{vmatrix} -2 & k \\ 1 & -1 \end{vmatrix} = 2 - k = 0$ .
Check $k = 2$ : trace 5, det $6 - 2 = 4$ , $\lambda^2 - 5\lambda + 4 = (\lambda - 1)(\lambda - 4)$ .

Verification: The characteristic polynomials were computed exactly in node with the Faddeev-LeVerrier method using BigInt fractions, and their roots were found with their multiplicities by a rational-root scan; each eigenspace basis was obtained as the exact null space of (A − λI), and the equality Av = λv was checked by multiplication for every eigenvector (verify/MAT1041/13.mjs).

Review card

Av = λv, v ≠ 0 → λ is an eigenvalue, v an eigenvector. (v = 0 does NOT count; λ = 0 is allowed.) Characteristic equation: det(λI - A) = 0 (or det(A - λI) = 0, same roots). 2×2 shortcut: λ² - (tr A)λ + det A = 0. Triangular/diagonal → eigenvalues on the diagonal. Eigenspace E_λ = null(A - λI): solve (A - λI)x = 0, write a basis with the free variables. (A - λI) must be SINGULAR; if you get only the trivial solution, you found λ wrongly. Algebraic multiplicity = multiplicity of the root. Geometric = dim E_λ. 1 ≤ geom ≤ algebraic. Σλ = tr A, Πλ = det A. λ = 0 is an eigenvalue ⇔ A is singular. A^k → λ^k; A⁻¹ → 1/λ; A + cI → λ + c; p(A) → p(λ). The eigenvector is the SAME. A and A^T have the same eigenvalues. Similar matrices have the same eigenvalues. There may be no real root (rotation: λ² + 1). For 3×3, expand det(λI - A) along the row with the most zeros, keep the factor.