8.3 Characteristic polynomials and diagonalization tests

The null-space criterion

\det(A-\lambda I)=0

tells you how to test a proposed eigenvalue. The next step is to package that test into one polynomial whose roots are exactly the eigenvalues of the matrix.

That polynomial is strong enough to limit how many eigenvalues a matrix can have, describe repeated eigenvalues, and give practical tests for diagonalization.

Characteristic polynomial

Definition

Characteristic polynomial

Let $A$ be an $n\times n$ matrix. The polynomial

p_A(x)=\det(A-xI_n)

is called the characteristic polynomial of $A$ .

The variable x is only an indeterminate. Once you substitute a scalar $\lambda$ , the value $p_A(\lambda)$ becomes the determinant of $A-\lambda I$ .

Theorem

Eigenvalues are exactly the roots of the characteristic polynomial

Let $A$ be an $n\times n$ matrix and $\lambda$ a scalar. Then the following are equivalent:

$\lambda$ is an eigenvalue of $A$ ;
$\lambda$ is a root of $p_A(x)$ .

This theorem is just the determinant test from the previous note written in a more organized way:

\lambda\text{ is an eigenvalue} \iff \det(A-\lambda I)=0 \iff p_A(\lambda)=0.

Theorem

Basic shape of the characteristic polynomial

If $A$ is an $n\times n$ matrix, then $p_A(x)$ is a degree-n polynomial whose leading coefficient is $(-1)^n$ and whose constant term is $\det(A)$ .

So the characteristic polynomial is never an arbitrary expression. Its degree is fixed by the matrix size, and its constant term already remembers the determinant.

First examples

Worked example

A 2×2 characteristic polynomial

Let

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

Then

p_A(x)= \det \begin{bmatrix} 3-x&2\\ 3&-2-x \end{bmatrix} =(3-x)(-2-x)-6.

Simplifying gives

p_A(x)=x^2-x-12=(x-4)(x+3).

Therefore the eigenvalues are 4 and $-3$ .

Worked example

A repeated root

Let

B= \begin{bmatrix} 2&3\\ 0&2 \end{bmatrix}.

Then

p_B(x)= \det \begin{bmatrix} 2-x&3\\ 0&2-x \end{bmatrix} =(2-x)^2=(x-2)^2.

So 2 is the only eigenvalue, but it appears with multiplicity 2.

Algebraic and geometric multiplicity

Repeated roots need extra language.

Definition

Algebraic multiplicity and geometric multiplicity

Suppose the characteristic polynomial factors as

p_A(x)=(-1)^n(x-\lambda_1)^{m_1}\cdots(x-\lambda_s)^{m_s},

where $\lambda_1,\dots,\lambda_s$ are distinct roots.

The exponent $m_i$ is the algebraic multiplicity of $\lambda_i$ .
The dimension of the eigenspace $E_A(\lambda_i)$ is the geometric multiplicity of $\lambda_i$ .

Algebraic multiplicity comes from the polynomial. Geometric multiplicity comes from the null space $N(A-\lambda I)$ . They measure different things, and they do not always agree.

Theorem

Multiplicity inequality

If $\lambda$ is an eigenvalue of an $n\times n$ matrix $A$ , then

1\le m_g(\lambda)\le m_a(\lambda)\le n.

So every eigenspace has dimension at least 1, but it can never be larger than the algebraic multiplicity of its eigenvalue.

Worked example

A repeated eigenvalue with too small an eigenspace

For

B= \begin{bmatrix} 2&3\\ 0&2 \end{bmatrix},

the characteristic polynomial is $(x-2)^2$ , so the eigenvalue 2 has algebraic multiplicity 2.

Now solve $(B-2I)x=0$ :

B-2I= \begin{bmatrix} 0&3\\ 0&0 \end{bmatrix}.

So $x_2=0$ and $x_1$ is free. The eigenspace is

\operatorname{span}\left\{ \begin{bmatrix} 1\\0 \end{bmatrix} \right\},

which has dimension 1. Therefore

m_a(2)=2,\qquad m_g(2)=1.

This mismatch is exactly why the matrix is not diagonalizable.

Distinct eigenvalues force independence

Theorem

Eigenvectors of distinct eigenvalues are linearly independent

If $v_1,\dots,v_k$ are eigenvectors of $A$ corresponding to pairwise distinct eigenvalues $\lambda_1,\dots,\lambda_k$ , then

v_1,\dots,v_k

are linearly independent.

This theorem has two major consequences.

Theorem

Upper bound on the number of distinct eigenvalues

An $n\times n$ matrix can have at most n distinct eigenvalues.

Theorem

Distinct eigenvalue test for diagonalizability

If an $n\times n$ matrix has exactly n distinct eigenvalues, then it is diagonalizable.

The converse is false. A diagonalizable matrix may still have repeated eigenvalues. The identity matrix is the simplest example: every vector is an eigenvector with eigenvalue 1, yet the matrix is already diagonal.

A sharper diagonalization criterion

Distinct eigenvalues are a sufficient test, but not the only one.

Theorem

Dimension-sum test for diagonalizability

Suppose a real $n\times n$ matrix $A$ has distinct real eigenvalues $\lambda_1,\dots,\lambda_s$ . Then $A$ is diagonalizable if and only if

\dim E_A(\lambda_1)+\cdots+\dim E_A(\lambda_s)=n.

This criterion says that diagonalization succeeds exactly when the eigenspaces together contribute enough independent eigenvectors to form a basis of the whole space.

Worked example

Repeated eigenvalue but still diagonalizable

Let

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

Its eigenvalues are 4 and 1, with 1 repeated algebraically. But the eigenspace for 4 is one-dimensional, while the eigenspace for 1 is two-dimensional. Therefore

1+2=3,

which matches the matrix size. So $A$ is diagonalizable even though one eigenvalue is repeated.

Cayley-Hamilton gives a polynomial identity

One further theorem relates the characteristic polynomial back to the matrix itself.

Theorem

Cayley-Hamilton theorem

p_A(x)=c_0+c_1x+c_2x^2+\cdots+c_nx^n,

then

c_0I+c_1A+c_2A^2+\cdots+c_nA^n=0.

Equivalently, $p_A(A)=0$ .

For diagonalizable matrices this is easy to believe: once $A$ is written as $SDS^{-1}$ , the identity reduces to applying the scalar polynomial $p_A$ to each diagonal entry of $D$ , and every diagonal entry is an eigenvalue, hence a root of $p_A$ .

The practical consequence is that high powers of $A$ can be rewritten in terms of lower powers $I,A,\dots,A^{n-1}$ .

Worked example

A small Cayley-Hamilton identity

Let

A= \begin{bmatrix} 1&4\\ 0&1 \end{bmatrix}.

Its characteristic polynomial is

p_A(x)=(1-x)^2=x^2-2x+1.

Therefore Cayley-Hamilton says

A^2-2A+I=0.

So every higher power of $A$ can be reduced using this quadratic relation.

Common mistake

Repeated eigenvalue does not automatically mean 'not diagonalizable'

A repeated eigenvalue only tells you the algebraic multiplicity is greater than 1. Diagonalizability depends on whether the corresponding eigenspaces still supply enough linearly independent eigenvectors. Repetition is a warning sign, not a final verdict.

Quick check

What is the characteristic polynomial of $\begin{bmatrix}2&0\\0&5\end{bmatrix}$ ?

Use $p_A(x)=\det(A-xI)$ .

Solution

Answer

Quick check

If an $n\times n$ matrix has n distinct eigenvalues, what can you conclude immediately?

Use the distinct-eigenvalue theorem.

Solution

Answer

Quick check

Can geometric multiplicity ever exceed algebraic multiplicity?

Use the multiplicity inequality.

Solution

Answer

Exercises

Quick check

Find the characteristic polynomial of $\begin{bmatrix}0&1\\-2&3\end{bmatrix}$ and list its eigenvalues.

Compute $\det(A-xI)$ and factor the result.

Solution

Guided solution

Quick check

A 4×4 matrix has four distinct eigenvalues. What is the dimension of the span of one chosen eigenvector from each eigenvalue?

Use linear independence of eigenvectors with distinct eigenvalues.

Solution

Guided solution

Quick check

Suppose $p_A(x)=x^3-6x^2+11x-6$ . What matrix identity does Cayley-Hamilton give?

Replace the variable x by the matrix $A$ .

Solution

Guided solution

Keep 8.1 Eigenvalues, eigenvectors, and eigenspaces open for the null-space interpretation of eigenvalues.

Keep 8.2 Diagonalization and similarity nearby, because this note supplies the polynomial tests that feed that chapter.

The determinant machinery inside $p_A(x)=\det(A-xI)$ depends on 7.1 Determinants and cofactor expansion.

8.3 Characteristic polynomials and diagonalization tests

MATH1030: Linear algebra I

Characteristic polynomial

Characteristic polynomial

Eigenvalues are exactly the roots of the characteristic polynomial

Basic shape of the characteristic polynomial

First examples

A 2×2 characteristic polynomial

A repeated root

Algebraic and geometric multiplicity

Algebraic multiplicity and geometric multiplicity

Multiplicity inequality

A repeated eigenvalue with too small an eigenspace

Distinct eigenvalues force independence

Eigenvectors of distinct eigenvalues are linearly independent

Upper bound on the number of distinct eigenvalues

Distinct eigenvalue test for diagonalizability

A sharper diagonalization criterion

Dimension-sum test for diagonalizability

Repeated eigenvalue but still diagonalizable

Cayley-Hamilton gives a polynomial identity

Cayley-Hamilton theorem

A small Cayley-Hamilton identity

Common mistake

Repeated eigenvalue does not automatically mean 'not diagonalizable'

Quick check

What is the characteristic polynomial of [2005]\begin{bmatrix}2&0\\0&5\end{bmatrix}[20​05​]?

Answer

If an n×nn\times nn×n matrix has n distinct eigenvalues, what can you conclude immediately?

Answer

Can geometric multiplicity ever exceed algebraic multiplicity?

Answer

Exercises

Find the characteristic polynomial of [01−23]\begin{bmatrix}0&1\\-2&3\end{bmatrix}[0−2​13​] and list its eigenvalues.

Guided solution

A 4×4 matrix has four distinct eigenvalues. What is the dimension of the span of one chosen eigenvector from each eigenvalue?

Guided solution

Suppose pA(x)=x3−6x2+11x−6p_A(x)=x^3-6x^2+11x-6pA​(x)=x3−6x2+11x−6. What matrix identity does Cayley-Hamilton give?

Guided solution

Related notes

Section mastery checkpoint

Prerequisites

Key terms in this unit

More notes in this series

What is the characteristic polynomial of $\begin{bmatrix}2&0\\0&5\end{bmatrix}$ ?

If an $n\times n$ matrix has n distinct eigenvalues, what can you conclude immediately?

Find the characteristic polynomial of $\begin{bmatrix}0&1\\-2&3\end{bmatrix}$ and list its eigenvalues.

Suppose $p_A(x)=x^3-6x^2+11x-6$ . What matrix identity does Cayley-Hamilton give?