Eigenvalues and eigenvectors

The characteristic equation, finding eigenvalues for 2×2 and 3×3 matrices, extracting eigenvectors, and the properties that let you check an answer in seconds.

Module contents

The idea

Multiplying a vector by a matrix normally changes both its direction and its length. For a few special directions, only the length changes. Those directions are the eigenvectors of the matrix, and the stretch factors are its eigenvalues.

$A X = λ X, X \neq = O$

Here $λ$ is a scalar (the eigenvalue) and $X$ a non-zero column vector (the eigenvector belonging to $λ$ ). The zero vector is excluded — it satisfies the equation for every $λ$ and would tell us nothing.

The characteristic equation

Rearrange:

$A X = λ X ⟹ A X - λ I X = O ⟹ (A - λ I) X = O$

Note $λ I$ , not $λ$ — you cannot subtract a scalar from a matrix.

This is a homogeneous system. It has a non-zero solution $X$ if and only if the coefficient matrix is singular:

$∣ A - λ I ∣ = 0$

This is the characteristic equation of $A$ . For a matrix of order $n$ it is a polynomial equation of degree $n$ in $λ$ , called the characteristic polynomial, and its $n$ roots are the eigenvalues. The set of all eigenvalues is the spectrum of $A$ .

Order 2

$∣ A - λ I ∣ = λ^{2} - (tr A) λ + ∣ A ∣ = 0$

Sum of the diagonal, then the determinant. Fast enough to do in your head.

Order 3

$λ^{3} - S_{1} λ^{2} + S_{2} λ - S_{3} = 0$

where

$S_{1} = tr A = a_{11} + a_{22} + a_{33}$
$S_{2} =$ sum of the three principal minors of order $2$ — the determinants left after deleting row $i$ and column $i$ , for $i = 1, 2, 3$
$S_{3} = ∣ A ∣$

Using this form is much quicker and safer than expanding $∣ A - λ I ∣$ symbolically.

Finding the eigenvectors

For each eigenvalue $λ_{i}$ in turn:

Form the matrix $A - λ_{i} I$ .
Solve $(A - λ_{i} I) X = O$ by row reduction.
Because the matrix is singular by construction, there is at least one free variable. Set it to a convenient value and read off the vector.

Eigenvectors are only determined up to a scalar multiple: if $X$ is an eigenvector then so is $c X$ for any $c \neq = 0$ . Choose whatever multiple keeps the entries as small integers.

The best check available. Once you have a candidate $X$ , compute $A X$ and confirm it equals $λ X$ . This catches every arithmetic error in the whole question, and takes seconds.

Properties worth memorising

For a matrix of order $n$ with eigenvalues $λ_{1}, λ_{2}, \dots, λ_{n}$ :

Property	Statement
Sum	$λ_{1} + λ_{2} + \dots + λ_{n} = tr A$
Product	$λ_{1} λ_{2} \dots λ_{n} = ∣ A ∣$
Transpose	$A$ and $A^{T}$ have the same eigenvalues
Scalar	eigenvalues of $k A$ are $k λ_{i}$
Powers	eigenvalues of $A^{m}$ are $λ_{i}^{m}$
Inverse	eigenvalues of $A^{- 1}$ are $1/ λ_{i}$ (needs all $λ_{i} \neq = 0$ )
Shift	eigenvalues of $A + k I$ are $λ_{i} + k$
Triangular	eigenvalues of a triangular or diagonal matrix are its diagonal entries
Singularity	$A$ is singular $⟺$ $0$ is an eigenvalue
Symmetric	a real symmetric matrix has all real eigenvalues
Orthogonal	eigenvalues have modulus $1$
Idempotent	eigenvalues are $0$ or $1$
Nilpotent	all eigenvalues are $0$

The sum and product checks are the fastest possible verification. Apply them before writing anything else down.

Where this is used

PageRank — the ranking vector is the dominant eigenvector of the web's link matrix.
Principal component analysis — the components are eigenvectors of the covariance matrix.
Vibration and stability analysis — eigenvalues are the natural frequencies; a positive real part means the system blows up.
Image compression and face recognition — via singular values, the close cousin of eigenvalues.

Worked examples

3 solved

Every step is shown, in the order you would write it in an answer book.

Example 1

Eigenvalues and eigenvectors of a 2 × 2

Question

Find the eigenvalues and corresponding eigenvectors of $A = [4213]$ .

Write the characteristic equation
$tr A = 4 + 3 = 7, ∣ A ∣ = 12 - 2 = 10$

$λ^{2} - 7 λ + 10 = 0$
Solve for $λ$
$(λ - 5) (λ - 2) = 0 ⟹ λ = 5, 2$

Sanity check: $5 + 2 = 7 = tr A$ ✓ and $5 \times 2 = 10 = ∣ A ∣$ ✓
Eigenvector for $λ = 5$
$A - 5 I = [- 1 2 1 - 2]$

Row 2 is $- 2 \times$ row 1, as it must be (the matrix is singular). The single independent equation is

$- x_{1} + x_{2} = 0 ⟹ x_{2} = x_{1}$

Taking $x_{1} = 1$ :

$X_{1} = [11]$
Eigenvector for $λ = 2$
$A - 2 I = [2211]$

The equation is $2 x_{1} + x_{2} = 0 \Rightarrow x_{2} = - 2 x_{1}$ .

Taking $x_{1} = 1$ :

$X_{2} = [1 - 2]$
Verify both
$A X_{1} = [4213] [11] = [55] = 5 [11] ✓$

$A X_{2} = [4213] [1 - 2] = [2 - 4] = 2 [1 - 2] ✓$

Answer

$λ = 5$ with $X_{1} = [11]$ ; $λ = 2$ with $X_{2} = [1 - 2]$ .

Example 2

A 3 × 3 with distinct eigenvalues

Question

Find the eigenvalues and eigenvectors of $A = 210120111$ .

Compute $S_{1}$ , $S_{2}$ , $S_{3}$
$S_{1} = tr A = 2 + 2 + 1 = 5$

Principal minors of order $2$ (delete row $i$ and column $i$ ):

$2011 = 2, 2011 = 2, 2112 = 3$

$S_{2} = 2 + 2 + 3 = 7$

$S_{3} = ∣ A ∣ = 2 (2 - 0) - 1 (1 - 0) + 1 (0 - 0) = 4 - 1 = 3$
Write and solve the characteristic equation
$λ^{3} - 5 λ^{2} + 7 λ - 3 = 0$

Try $λ = 1$ : $1 - 5 + 7 - 3 = 0$ ✓, so $(λ - 1)$ is a factor.

$λ^{3} - 5 λ^{2} + 7 λ - 3 = (λ - 1) (λ^{2} - 4 λ + 3) = (λ - 1) (λ - 1) (λ - 3)$

$λ = 1, 1, 3$

Check: $1 + 1 + 3 = 5 = S_{1}$ ✓ and $1 \times 1 \times 3 = 3 = ∣ A ∣$ ✓
Eigenvector for $λ = 3$
$A - 3 I = - 1 10 1 - 1 0 11 - 2$

Row 3: $- 2 x_{3} = 0 \Rightarrow x_{3} = 0$ .

Row 1: $- x_{1} + x_{2} + 0 = 0 \Rightarrow x_{2} = x_{1}$ .

Taking $x_{1} = 1$ :

$X_{1} = 110$

Check: $A X_{1} = (2 + 1 + 0, 1 + 2 + 0, 0) = (3, 3, 0) = 3 X_{1}$ ✓
Eigenvectors for the repeated $λ = 1$
$A - I = 110110110$

All rows reduce to the single equation

$x_{1} + x_{2} + x_{3} = 0$

Rank is $1$ , so $n - ρ = 3 - 1 = 2$ free variables — this repeated eigenvalue supplies two independent eigenvectors.

Choose $x_{2} = 1, x_{3} = 0 \Rightarrow x_{1} = - 1$ :
$X_{2} = - 1 10$

Choose $x_{2} = 0, x_{3} = 1 \Rightarrow x_{1} = - 1$ :
$X_{3} = - 1 01$

Check $X_{2}$ : $A X_{2} = (- 2 + 1 + 0, - 1 + 2 + 0, 0) = (- 1, 1, 0) = 1 \cdot X_{2}$ ✓

Answer

$λ = 3$ : $X_{1} = 110$ . $λ = 1$ (twice): $X_{2} = - 1 10$ , $X_{3} = - 1 01$ .

NoteA repeated eigenvalue does not always give two independent eigenvectors. Count them by computing the rank of A − λI: the number is n − ρ(A − λI).

Example 3

Reading eigenvalues off a triangular matrix

Question

Find the eigenvalues of $A = 300 7 - 1 0 - 2 54$ and hence the eigenvalues of $A^{2}$ , $A^{- 1}$ and $A + 2 I$ .

Use the triangular shortcut
$A$ is upper triangular, so $∣ A - λ I ∣$ is also triangular and equals

$(3 - λ) (- 1 - λ) (4 - λ) = 0$

$λ = 3, - 1, 4$

No expansion needed.
Confirm with the trace and determinant
$tr A = 3 - 1 + 4 = 6$ and $3 + (- 1) + 4 = 6$ ✓

$∣ A ∣ = 3 (- 1) (4) = - 12$ and $3 \times (- 1) \times 4 = - 12$ ✓
Apply the property rules
$A^{2}$ : square each eigenvalue — $9, 1, 16$

$A^{- 1}$ : reciprocate each — $\frac{1}{3}, - 1, \frac{1}{4}$ (valid since no eigenvalue is $0$ )

$A + 2 I$ : add $2$ to each — $5, 1, 6$

Answer

$A$ : $3, - 1, 4$ . $A^{2}$ : $9, 1, 16$ . $A^{- 1}$ : $1/3, - 1, 1/4$ . $A + 2 I$ : $5, 1, 6$ .

NoteQuestions about A², A⁻¹ or A + kI never require recomputing a characteristic equation. Transform the eigenvalues you already have.

Practice problems

6 with solutions

Work each one on paper first. The full solution — not just the answer — is one click away.

Problem 1Basic

Find the eigenvalues of $A = [5142]$ .

Show solution

$tr A = 5 + 2 = 7, ∣ A ∣ = 10 - 4 = 6$

$λ^{2} - 7 λ + 6 = 0 ⟹ (λ - 6) (λ - 1) = 0$

$λ = 6, 1$

Check: $6 + 1 = 7$ ✓, $6 \times 1 = 6$ ✓

Answer

λ = 6, 1

Problem 2Basic

Find the eigenvalues and eigenvectors of $A = [1322]$ .

Show solution

$tr A = 3, ∣ A ∣ = 2 - 6 = - 4$

$λ^{2} - 3 λ - 4 = 0 ⟹ (λ - 4) (λ + 1) = 0 ⟹ λ = 4, - 1$

For $λ = 4$ :
$A - 4 I = [- 3 3 2 - 2] ⟹ - 3 x_{1} + 2 x_{2} = 0 ⟹ x_{2} = \frac{3}{2} x_{1}$

Take $x_{1} = 2$ : $X_{1} = [23]$

For $λ = - 1$ :
$A + I = [2323] ⟹ x_{1} + x_{2} = 0 ⟹ x_{2} = - x_{1}$

Take $x_{1} = 1$ : $X_{2} = [1 - 1]$

Verify: $A X_{1} = (2 + 6, 6 + 6) = (8, 12) = 4 X_{1}$ ✓ $A X_{2} = (1 - 2, 3 - 2) = (- 1, 1) = - X_{2}$ ✓

Answer

λ = 4

with

[23]

;

λ = - 1

with

[1 - 1]

Problem 3Exam level

Find the characteristic equation and eigenvalues of $A = 113151311$ .

Show solution

$S_{1} = 1 + 5 + 1 = 7$

Principal $2 \times 2$ minors:

$5111 = 4, 1331 = - 8, 1115 = 4$

$S_{2} = 4 - 8 + 4 = 0$

$S_{3} = ∣ A ∣ = 1 (5 - 1) - 1 (1 - 3) + 3 (1 - 15) = 4 + 2 - 42 = - 36$

Characteristic equation:

$λ^{3} - 7 λ^{2} + 0 λ + 36 = 0 ⟹ λ^{3} - 7 λ^{2} + 36 = 0$

Try $λ = - 2$ : $- 8 - 28 + 36 = 0$ ✓

Divide: $λ^{3} - 7 λ^{2} + 36 = (λ + 2) (λ^{2} - 9 λ + 18) = (λ + 2) (λ - 3) (λ - 6)$

$λ = - 2, 3, 6$

Check: $- 2 + 3 + 6 = 7 = S_{1}$ ✓ $(- 2) (3) (6) = - 36 = ∣ A ∣$ ✓

(All eigenvalues are real, as they must be — $A$ is symmetric.)

Answer

λ^{3} - 7 λ^{2} + 36 = 0

;

λ = - 2, 3, 6

Problem 4Exam level

Find the eigenvalues and eigenvectors of $A = 211232112$ .

Show solution

Characteristic equation.

$S_{1} = 2 + 3 + 2 = 7$

$S_{2} = 3212 + 2112 + 2123 = 4 + 3 + 4 = 11$

$S_{3} = ∣ A ∣ = 2 (6 - 2) - 2 (2 - 1) + 1 (2 - 3) = 8 - 2 - 1 = 5$

$λ^{3} - 7 λ^{2} + 11 λ - 5 = 0$

Try $λ = 1$ : $1 - 7 + 11 - 5 = 0$ ✓

$(λ - 1) (λ^{2} - 6 λ + 5) = (λ - 1) (λ - 1) (λ - 5)$

$λ = 1, 1, 5$

Eigenvector for $λ = 5$ .

$A - 5 I = - 3 11 2 - 2 2 11 - 3$

$R_{1} \leftrightarrow R_{2}$ , then $R_{2} \to R_{2} + 3 R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :

$100 - 2 - 4 4 14 - 4 \sim 100 - 2 - 4 0 140$

Row 2: $- 4 x_{2} + 4 x_{3} = 0 \Rightarrow x_{2} = x_{3}$ .
Row 1: $x_{1} - 2 x_{2} + x_{3} = 0 \Rightarrow x_{1} = 2 x_{2} - x_{3} = x_{3}$ .

Take $x_{3} = 1$ : $X_{1} = 111$

Check: $A X_{1} = (2 + 2 + 1, 1 + 3 + 1, 1 + 2 + 2) = (5, 5, 5) = 5 X_{1}$ ✓

Eigenvectors for $λ = 1$ .

$A - I = 111222111$

Rank $1$ , so $3 - 1 = 2$ independent eigenvectors. The single equation is

$x_{1} + 2 x_{2} + x_{3} = 0$

Take $x_{2} = 1, x_{3} = 0 \Rightarrow x_{1} = - 2$ : $X_{2} = - 2 10$

Take $x_{2} = 0, x_{3} = 1 \Rightarrow x_{1} = - 1$ : $X_{3} = - 1 01$

Check $X_{2}$ : $A X_{2} = (- 4 + 2 + 0, - 2 + 3 + 0, - 2 + 2 + 0) = (- 2, 1, 0) = X_{2}$ ✓

Answer

λ = 5

111

;

λ = 1

(twice):

- 2 10

and

- 1 01

Problem 5Exam level

If the eigenvalues of a $3 \times 3$ matrix $A$ are $2, - 3, 4$ , find $tr A$ , $∣ A ∣$ , the eigenvalues of $A^{3}$ , of $A^{- 1}$ , and of $3 A - I$ .

Show solution

Trace = sum of eigenvalues:
$tr A = 2 + (- 3) + 4 = 3$

Determinant = product of eigenvalues:
$∣ A ∣ = 2 \times (- 3) \times 4 = - 24$

Non-zero, so $A^{- 1}$ exists.

$A^{3}$ — cube each eigenvalue:
$8, - 27, 64$

$A^{- 1}$ — reciprocate each:
$\frac{1}{2}, - \frac{1}{3}, \frac{1}{4}$

$3 A - I$ — apply $λ \mapsto 3 λ - 1$ :
$3 (2) - 1 = 5, 3 (- 3) - 1 = - 10, 3 (4) - 1 = 11$

(In general, if $p$ is any polynomial then the eigenvalues of $p (A)$ are $p (λ_{i})$ .)

Answer

tr A = 3

∣ A ∣ = - 24

;

A^{3}

8, - 27, 64

;

A^{- 1}

1/2, - 1/3, 1/4

;

3 A - I

5, - 10, 11

Problem 6Challenge

Prove that if $λ$ is an eigenvalue of a non-singular matrix $A$ , then $1/ λ$ is an eigenvalue of $A^{- 1}$ with the same eigenvector.

Show solution

Start from the definition, with $X \neq = O$ :

$A X = λ X$

First, $λ \neq = 0$ . If $λ$ were $0$ then $A X = O$ with $X \neq = O$ , which would make $A$ singular — contradicting the hypothesis. So dividing by $λ$ is legal.

Multiply both sides on the left by $A^{- 1}$ :

$A^{- 1} (A X) = A^{- 1} (λ X)$

$(A^{- 1} A) X = λ A^{- 1} X$

$I X = λ A^{- 1} X ⟹ X = λ A^{- 1} X$

Divide by $λ$ :

$A^{- 1} X = \frac{1}{λ} X$

This is exactly the eigenvalue equation for $A^{- 1}$ , with eigenvalue $1/ λ$ and the same eigenvector $X$ . ∎

The same argument applied repeatedly gives the general rule: $A^{m} X = λ^{m} X$ for every integer $m$ , positive or negative.

Answer

A X = λ X \Rightarrow A^{- 1} X = (1/ λ) X

, same eigenvector.