Math Core

In the last lesson you could test whether a given number is an eigenvalue. But how do you find the eigenvalues when nobody hands you a candidate? Determinants answer that question: they turn " $A - \lambda I$ is not invertible" into a polynomial equation in $\lambda$ whose roots are exactly the eigenvalues.

From invertibility to a polynomial

A scalar $\lambda$ is an eigenvalue of $A$ exactly when $(A - \lambda I)\mathbf{x} = \mathbf{0}$ has a nontrivial solution, which by the Invertible Matrix Theorem happens exactly when $A - \lambda I$ is not invertible, which happens exactly when its determinant is zero.

The characteristic equation

A scalar $\lambda$ is an eigenvalue of the $n \times n$ matrix $A$ if and only if it satisfies the characteristic equation

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

The expression $\det(A - \lambda I)$ is a polynomial of degree $n$ in $\lambda$ , called the characteristic polynomial of $A$ .

Why is it a polynomial of degree $n$ ? Each entry of $A - \lambda I$ is either a constant or a constant minus $\lambda$ , and the determinant is a sum of products of $n$ entries. The product of the diagonal entries contributes $(-\lambda)^n$ , and no other term has that many factors of $\lambda$ .

The $2 \times 2$ case

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ ,

\det(A - \lambda I) = (a - \lambda)(d - \lambda) - bc = \lambda^2 - (a + d)\lambda + (ad - bc).

The coefficient $a + d$ is the trace of $A$ (the sum of the diagonal entries), and the constant term is $\det A$ . So every $2 \times 2$ characteristic polynomial is

\lambda^2 - (\operatorname{tr} A)\,\lambda + \det A.

This is worth memorizing; it saves time on every $2 \times 2$ problem.

Worked example: Integer eigenvalues

Find the eigenvalues of $A = \begin{bmatrix} 2 & 3 \\ 3 & -6 \end{bmatrix}$ .

Here $\operatorname{tr} A = -4$ and $\det A = -12 - 9 = -21$ , so the characteristic equation is

\lambda^2 + 4\lambda - 21 = (\lambda + 7)(\lambda - 3) = 0.

The eigenvalues are $3$ and $-7$ .

Worked example: Irrational eigenvalues

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

$\operatorname{tr} A = 4$ and $\det A = 3 - 2 = 1$ , so $\lambda^2 - 4\lambda + 1 = 0$ . The quadratic formula gives

\lambda = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}.

Eigenvalues do not have to be nice numbers, and (as the lesson on complex eigenvalues will show) they do not even have to be real.

Larger matrices and multiplicity

For a $3 \times 3$ matrix you compute $\det(A - \lambda I)$ by cofactor expansion, choosing a row or column with zeros to keep the algebra short. Try to leave the answer factored as long as you can; multiplying everything out and then factoring a cubic is much harder.

Worked example: A 3 × 3 matrix

Find the characteristic polynomial and eigenvalues of $A = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & -1 \\ 4 & 2 & 0 \end{bmatrix}$ .

Expand $\det(A - \lambda I)$ along the first row, which has two zeros:

\begin{aligned} \det(A - \lambda I) &= (2 - \lambda)\det\begin{bmatrix} 3 - \lambda & -1 \\ 2 & -\lambda \end{bmatrix} \\ &= (2 - \lambda)\big[(3 - \lambda)(-\lambda) + 2\big] \\ &= (2 - \lambda)(\lambda^2 - 3\lambda + 2) \\ &= (2 - \lambda)(\lambda - 1)(\lambda - 2) = -(\lambda - 2)^2(\lambda - 1). \end{aligned}

The eigenvalues are $2$ and $1$ , and the factor $\lambda - 2$ appears twice.

Definition

Algebraic multiplicity

The algebraic multiplicity of an eigenvalue $\lambda_0$ is the number of times $(\lambda - \lambda_0)$ appears as a factor of the characteristic polynomial.

In the example, $\lambda = 2$ has algebraic multiplicity $2$ and $\lambda = 1$ has multiplicity $1$ . Counting with multiplicity (and allowing complex roots), an $n \times n$ matrix always has exactly $n$ eigenvalues, because a degree- $n$ polynomial has $n$ roots.

Two consequences follow from comparing coefficients. If the eigenvalues of $A$ , listed with multiplicity, are $\lambda_1, \dots, \lambda_n$ , then

\det A = \lambda_1\lambda_2\cdots\lambda_n \qquad\text{and}\qquad \operatorname{tr} A = \lambda_1 + \lambda_2 + \cdots + \lambda_n.

(The determinant fact comes from setting $\lambda = 0$ in $\det(A - \lambda I) = (\lambda_1 - \lambda)\cdots(\lambda_n - \lambda)$ .) These give quick checks: in the $3 \times 3$ example, $2 \cdot 2 \cdot 1 = 4 = \det A$ and $2 + 2 + 1 = 5 = \operatorname{tr} A$ .

Tip

After finding eigenvalues, check that they add up to the trace and multiply to the determinant. It catches most arithmetic slips in seconds.

Similar matrices

Definition

Similarity

Square matrices $A$ and $B$ are similar if there is an invertible matrix $P$ with $B = P^{-1}AP$ (equivalently, $A = PBP^{-1}$ ). Changing $A$ into $P^{-1}AP$ is called a similarity transformation.

Similar matrices describe the same linear transformation in different coordinate systems, just as in the change-of-basis lesson. So it is no surprise that they share their eigenvalues. In fact they share the whole characteristic polynomial:

B - \lambda I = P^{-1}AP - \lambda P^{-1}P = P^{-1}(A - \lambda I)P,

so by the multiplicative property of determinants,

\det(B - \lambda I) = \det(P^{-1})\det(A - \lambda I)\det(P) = \det(A - \lambda I).

Common mistake

Two cautions. First, the converse is false: $\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ and $\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ have the same characteristic polynomial, $(\lambda - 2)^2$ , but are not similar (the only matrix similar to $2I$ is $2I$ itself). Second, row equivalence is not similarity. Row operations usually change the eigenvalues, so never read eigenvalues off an echelon form of $A$ .

Practice

Practice 1

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

Separate answers with commas, e.g. 2, -5

Practice 2

The characteristic polynomial of $\begin{bmatrix} 7 & 2 \\ -3 & 1 \end{bmatrix}$ is $\lambda^2 + b\lambda + c$ . Find $c$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find the eigenvalues of $\begin{bmatrix} 4 & 1 \\ 2 & 2 \end{bmatrix}$ . Give exact values.

Separate answers with commas, e.g. 2, -5

Practice 4

Find the algebraic multiplicity of the eigenvalue $\lambda = 1$ for $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 4 & 3 \\ -1 & 1 & 2 \end{bmatrix}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Find the value of $h$ for which $A = \begin{bmatrix} 3 & h \\ 2 & 4 \end{bmatrix}$ has a repeated eigenvalue.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

A $3 \times 3$ matrix $A$ has eigenvalues $2$ , $-1$ and $4$ . Find $\det A$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Which statement is always true for $n \times n$ matrices?

Practice 8

Find all eigenvalues of $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{bmatrix}$ .

Separate answers with commas, e.g. 2, -5

From invertibility to a polynomial

The 2×22 \times 22×2 case

Larger matrices and multiplicity

Similar matrices

Practice

The $2 \times 2$ case