Skip to main content

Section6.2The Characteristic Polynomial

Objectives
  1. Learn that the eigenvalues of a triangular matrix are the diagonal entries.
  2. Learn some strategies for finding the zeros of a polynomial.
  3. Recipes: compute eigenvalues using the characteristic polynomial, the characteristic polynomial of a 2 × 2 matrix.
  4. Vocabulary words: characteristic polynomial, trace.

In Section 6.1 we discussed how to decide whether a given number λ is an eigenvalue of a matrix, and if so, how to find all of the associated eigenvectors. In this section, we will give a method for computing all of the eigenvalues of a matrix. This does not reduce to solving a system of linear equations: indeed, it requires solving a nonlinear equation in one variable, namely, finding the roots of the characteristic polynomial.

Definition

Let A be an n × n matrix. The characteristic polynomial of A is the function

f ( λ )= det ( A λ I n ) .

We will see below that the characteristic polynomial is in fact a polynomial. Finding the characterestic polynomial means computing the determinant of the matrix A λ I n , whose entries contain the unknown λ .

The point of the characteristic polynomial is that we can use it to compute eigenvalues.

Proof

By the invertible matrix theorem in Section 6.1, the matrix equation ( A λ I n ) x = 0 has a nontrivial solution if and only if det ( A λ I n )= 0. Therefore,

λ isaneigenvalueof A ⇐⇒ Ax = λ x hasanontrivialsolution ⇐⇒ ( A λ I n ) x = 0hasanontrivialsolution ⇐⇒ A λ I n isnotinvertible ⇐⇒ det ( A λ I n )= 0 ⇐⇒ f ( λ )= 0.
Form of the characteristic polynomial

It is time that we justified the use of the term “polynomial.” First we need a vocabulary word.

Definition

The trace of a square matrix A is the number Tr ( A ) obtained by summing the diagonal entries of A :

Tr EIIIIG a 11 a 12 ··· a 1, n 1 a 1 n a 21 a 22 ··· a 2, n 1 a 2 n ............... a n 1,1 a n 1,2 ··· a n 1, n 1 a n 1, n a n 1 a n 2 ··· a n , n 1 a nn FJJJJH = a 11 + a 22 + ··· + a nn .
Proof

First we notice that

f ( 0 )= det ( A 0 I n )= det ( A ) ,

so that the constant term is always det ( A ) .

We will prove the rest of the theorem only for 2 × 2 matrices; the reader is encouraged to complete the proof in general using cofactor expansions. We can write a 2 × 2 matrix as A = A abcd B ; then

f ( λ )= det ( A λ I 2 )= det K a λ bcd λ L =( a λ )( d λ ) bc = λ 2 ( a + d ) λ +( ad bc )= λ 2 Tr ( A ) λ + det ( A ) .
Recipe: The characteristic polynomial of a 2 × 2 matrix

When n = 2, the theorem tells us all of the coefficients of the characteristic polynomial:

f ( λ )= λ 2 Tr ( A ) λ + det ( A ) .

This is generally the fastest way to compute the characteristic polynomial of a 2 × 2 matrix.

Factoring the characteristic polynomial

If A is an n × n matrix, then the characteristic polynomial f ( λ ) has degree n by the above theorem. When n = 2, one can use the quadratic formula to find the roots of f ( λ ) . There exist algebraic formulas for the roots of cubic and quartic polynomials, but these are generally too cumbersome to apply by hand. Even worse, it is known that there is no algebraic formula for the roots of a general polynomial of degree at least 5.

In practice, the roots of the characteristic polynomial are found numerically by computer. That said, there do exist methods for finding roots by hand. For instance, we have the following consequence of the rational root theorem (which we also call the rational root theorem):

For example, if A has integer entries, then its characteristic polynomial has integer coefficients. This gives us one way to find a root by hand, if A has an eigenvalue that is a rational number. Once we have found one root, then we can reduce the degree by polynomial long division.

In the above example, we could have expanded cofactors along the second column to obtain

f ( λ )=( 2 λ ) det K 7 λ 3 3 1 λ L .

Since 2 λ was the only nonzero entry in its column, this expression already has the 2 λ term factored out: the rational root theorem was not needed. The determinant in the above expression is the characteristic polynomial of the matrix A 73 3 1 B , so we can compute it using the trace and determinant:

f ( λ )=( 2 λ ) A λ 2 ( 7 1 ) λ +( 7 + 9 ) B =( 2 λ )( λ 2 6 λ + 2 ) .
Finding Eigenvalues of a 3 × 3 Matrix

Let A be a 3 × 3 matrix. Here are some strategies for factoring its characteristic polynomial f ( λ ) . First, you must find one eigenvalue:

  1. Do not multiply out the characteristic polynomial if it is already partially factored! This happens if you expand cofactors along the second column in this example.
  2. If there is no constant term, you can factor out λ , as in this example.
  3. If the matrix is triangular, the roots are the diagonal entries (see below).
  4. Guess one eigenvalue using the rational root theorem: if det ( A ) is an integer, substitute all (positive and negative) divisors of det ( A ) into f ( λ ) .
  5. Find an eigenvalue using the geometry of the matrix. For instance, a reflection has eigenvalues ± 1.

After obtaining an eigenvalue λ 1 , use polynomial long division to compute f ( λ ) / ( λ λ 1 ) . This is a quadratic polynomial, to which you can apply the quadratic formula to find the remaining roots.

Eigenvalues of a triangular matrix

It is easy to compute the determinant of an upper- or lower-triangular matrix; this makes it easy to find its eigenvalues as well.

Proof

Suppose for simplicity that A is a 3 × 3 upper-triangular matrix:

A = C a 11 a 12 a 13 0 a 22 a 23 00 a 33 D .

Its characteristic polynomial is

f ( λ )= det ( A λ I 3 )= det C a 11 λ a 12 a 13 0 a 22 λ a 23 00 a 33 λ D .

This is also an upper-triangular matrix, so the determinant is the product of the diagonal entries:

f ( λ )=( a 11 λ )( a 22 λ )( a 33 λ ) .

The zeros of this polynomial are exactly a 11 , a 22 , a 33 .