6.1 Eigenvalues and Eigenvectors · Linear Algebra Refresher

Most directions get rotated and stretched by a matrix. Eigenvectors are the special directions that only get scaled. Find them and you've found the coordinate system in which the matrix is as simple as it can possibly be.

Definition · Eigenvalue and eigenvector

Let

A

n \times n

. A scalar

λ

is an eigenvalue of

A

if there exists a nonzero vector

x

with

A x = λ x

; such an

x

is an eigenvector belonging to

λ

Rewriting $A x = λ x$ as $(A - λ I) x = 0$ turns the problem into a familiar one: $λ$ is an eigenvalue exactly when this homogeneous system has a nontrivial solution, i.e. when $A - λ I$ is singular.

Equivalent conditions

For an

n \times n

matrix

A

and scalar

λ

, the following are equivalent:

λ

is an eigenvalue of

A

;

(A - λ I) x = 0

has a nontrivial solution;

N (A - λ I) \neq = {0}

;

A - λ I

is singular;

det (A - λ I) = 0

Definition · Characteristic polynomial, eigenspace

p (λ) = det (A - λ I)

is the characteristic polynomial (degree

n

), and

p (λ) = 0

is the characteristic equation. The subspace

N (A - λ I)

is the eigenspace of

λ

; its dimension is the geometric multiplicity, and the multiplicity of

λ

as a root of

p

is the algebraic multiplicity.

!Exam trap

Eigenvectors must be nonzero — otherwise every scalar would qualify. But

λ = 0

is allowed, and it occurs exactly when

A

is singular (then the eigenspace is

N (A)

). Don't confuse 'zero eigenvalue' with 'zero eigenvector'.

Finding eigenvalues and eigenspaces

1
Form $A - λ I$ — subtract $λ$ from the diagonal only
2
Compute $det (A - λ I)$ and set it to zero
Use §2.1 tactics: expand along the zero-richest row or column. For $2 \times 2$ there's a shortcut using trace and determinant.
$2 \times 2 : p (λ) = λ^{2} - tr (A) λ + det (A)$
3
Factor to get the eigenvalues with multiplicity
Try small integer roots first — exam matrices are built to factor. Verify with $\sum λ_{i} = tr (A)$ and $\prod λ_{i} = det (A)$ .
4
For each $λ$ , row reduce $A - λ I$ and solve $(A - λ I) x = 0$
There must be at least one free variable — if there isn't, $λ$ is wrong or the arithmetic slipped.
5
Write a basis for each eigenspace
Set each free variable to 1 in turn. Scale to clear fractions — any nonzero multiple of an eigenvector is an eigenvector.
6
Compare geometric and algebraic multiplicities
equal for all λ $n$ independent eigenvectors exist ⟹ diagonalizable (§6.3).
some geometric < algebraicDefective — not diagonalizable.

Every eigenvalue problem, every time. Steps 4–5 are just §3.2's null-space routine.

Worked example

A repeated eigenvalue that still behaves

0/5 steps

Find the eigenvalues and a basis for each eigenspace.

A = 211 - 3 - 2 - 3 112

Complex eigenvalues, trace, determinant, similarity

A real matrix has a real characteristic polynomial, so complex roots come in conjugate pairs: if $λ = a + bi$ is an eigenvalue with eigenvector $z$ , then $λ ˉ = a - bi$ is an eigenvalue with eigenvector $z ˉ$ , because $A z ˉ = A z = λ z = λ ˉ z ˉ$ .

Trace and determinant

λ_{1}, \dots, λ_{n}

are the eigenvalues of

A

counted with multiplicity, then

det (A) = i = 1 \prod n λ_{i}, tr (A) = i = 1 \sum n a_{ii} = i = 1 \sum n λ_{i} .

Use these constantly. If your eigenvalues don't sum to the trace, you've made an error — no need to check anything else.

Definition · Similar matrices

B

is similar to

A

B = S^{- 1} A S

for some nonsingular

S

. From §4.2: similar matrices represent the same linear operator in different bases.

Theorem 6.1.1

Similar matrices have the same characteristic polynomial, hence the same eigenvalues (with the same algebraic multiplicities), the same trace, and the same determinant.

Proof: $det (S^{- 1} A S - λ I) = det (S^{- 1} (A - λ I) S) = det (S^{- 1}) det (A - λ I) det (S) = det (A - λ I)$ . But the eigenvectors transform: $x \mapsto S^{- 1} x$ .

Check your work

Eigenvalues, eigenspaces & diagonalizability

Builds the characteristic polynomial exactly, factors out rational roots, then solves the null space of A − λI.

Square matrix

characteristic polynomial

p(λ) = λ³ − 2λ² + λ = 0

trace = Σλ

det = Πλ

diagonalizable

yes

λ = 0algebraic mult. 1 · geometric mult. 1

eigenspace basis:

λ = 1algebraic mult. 2 · geometric mult. 2

eigenspace basis:

-1

Check yourself

T / F

0

is an eigenvector of every square matrix, since

A 0 = λ 0

for any

λ

λ = 0

is an eigenvalue of

A

if and only if:

3 \times 3

matrix has eigenvalues

2

- 1

, and

4

. What are

tr (A)

and

det (A)

T / F

If a real

4 \times 4

matrix has

2 + 3 i

as an eigenvalue, then

2 - 3 i

is also an eigenvalue.

T / F

Similar matrices have the same eigenvalues and the same eigenvectors.

6.1Eigenvalues and Eigenvectors

What you must be able to do

Finding eigenvalues and eigenspaces

A repeated eigenvalue that still behaves

Complex eigenvalues, trace, determinant, similarity

Eigenvalues, eigenspaces & diagonalizability

Check yourself