### Sec 4.4

```Chapter 4
Section 4.4
Eigenvalues and the Characteristic
Polynomial
Characteristic Polynomial
If A is an  ×  matrix the characteristic polynomial is a function
of the variable t we call   that is the determinant of  − .
= det  −
Characteristic Equation
If   is the characteristic polynomial of a matrix A, setting   = 0 is the characteristic
equation of the matrix A. The solutions to this equation are the eigenvalues of matrix A.
Example
For the matrix A given to the right find its characteristic
polynomial and its eigenvalues.
9
5 2
= −10 −6 1
0
0 3
Since the 3rd row has two zeros we compute det  −  by expanding across the third row.
9−
−10
0
5
−6 −
0
2
9−
1 = 3−
−10
3−
5
= 3−
−6 −
9 −  −6 −  − −50
= 3 −   2 − 3 − 54 + 50 = 3 −   2 − 3 − 4 = 3 −   − 4  + 1
Since the object of these problems is to find the roots of this polynomial we very often
will leave the characteristic polynomial in factored form and do not multiply it out so we
can tell what the solutions to   = 0 are by just looking at it.
Characteristic Equation:   = 3 −   − 4  + 1
Eigenvalues are: -1, 3, 4
Example
Find the characteristic polynomial and the eigenvalues for the matrix
A given to the right. (Sometimes called Fibonacci matrix.)
Characteristic Polynomial is:   = det  −  = −
1
=
0
1
1
1
1
= − 1 −  − 1 =  2 −  − 1
1−
This polynomial does not factor so use the quadratic formula to get  =
1± 12 − −4
2
=
1± 5
2
If A is a  ×  matrix and   = det  −  is the characteristic polynomial for A then
is a polynomial of degree n with the coefficient of   being ±1. Using algebraic methods to
solve a general polynomial of degree n can be difficult (in fact, sometimes impossible), but
there are some techniques that will often work.
Factoring and Solving Polynomials
If   =   + −1  −1 + ⋯ + 1  + 0 is a polynomial with integer coefficients sometimes
the following method will work for factoring the polynomial   .
1. Find the factors of the constant term 0 say they are ±1, ±1 , ⋯ ±  .
2. Plug each factor into   till you find one that is zero say  (i.e.   = 0), that means
that  −  is one of the factors.
3. Use long division to find   ÷  −  =   that means that   =  −    .
This gives one factor of   .
4. Apply the same thing to   until you get down to degree 1 or 2 polynomial.
Example
12
4
6
Find the characteristic polynomial and eigenvalues for the
= −19 −6 −11
matrix A given to the right.
−9 −4 −3
12 −
4
6
= det  −  = −19 −6 −
−11
−9
−4
−3 −
Use a 1st
−6 −
−11
−19
−11
−19 −6 −
column
= 12 −
−4
+6
−4
−3 −
−9 −3 −
−9
−4
expansion
= 12 −  −6 −  −3 −  − 44 − 4 19 + 57 − 99 + 6 76 − 54 + 9
then
2
= 12 −   + 9 + 18 − 44 − 4 19 − 42 + 6 −9 + 22
simplify
2
= 12 −   + 9 − 26 − 76 + 168 − 54 + 132
the
2
3
2
polynomial
= 12 + 108 − 312 −  − 9 + 26 − 130 + 300
= − 3 + 3 2 + 4 − 12
= −  3 − 3 2 − 4 + 12
1. The constant term is 12, and the factors of 12 are:
± 1, ±2, ±3, ±4, ±6, ±12
2. Plugging in we get  1 = −6,  −1 = −12,  2 = 0
2 −  − 6
−2
3 −3 2 −4+12
Use long division to get
−2
Now we know   = −  − 2  2 −
3.
4.
−6
5. Factor   = −  − 2  + 2  − 3
The characteristic polynomial is:
= − 3 + 3 2 + 4 − 12
Eigenvalues are: −2,2,3
3 − 3 2 − 4 + 12
−  3 − 2 2
− 2 − 4 + 12
− − 2 + 2
−6 + 12
− −6 + 12
0
In general the problem of finding the characteristic polynomial and corresponding eigenvalues
for a matrix can be very difficult. If the matrix has a certain pattern or contains a lot of zeros
there are quick and easy ways to find the characteristic polynomial and eigenvalues.
Characteristic Polynomials for Diagonal, Lower and Upper Triangular Matrices
If  =  diagonal or triangular:
= 11 −  22 −  ⋯  −
The characteristic polynomial of a diagonal,
upper or lower triangular matrix  =  is
the product of each factor  −  and the
eigenvalues are the main diagonal entries.
The Eigenvalues are: 11 , 22 , ⋯ ,
Example
Find the characteristic polynomial and eigenvalues for the matrix A below to the right.
0 0 0 0
5
The matrix is upper triangular.
6 0 0 0
3
= 5 −  8 −  −1 −  6 −  2
= −1 1 8 0 0
0 −2 4 −1 0
Eigenvalues: −1, 5, 6, 8
7 0 9 6
6
Example
Find three different 3 × 3 upper triangular matrices whose only eigenvalues are 1 and 4.
Give all the possibilities for the characteristic polynomial
1 ∗ ∗
1 ∗ ∗
1 ∗ ∗
= 0 4 ∗
= 0 1 ∗
= 0 4 ∗
0 0 1
0 0 4
0 0 4
Characteristic polynomial   = −  − 1
2
− 4 or   = −  − 1  − 4
2
The number of times an eigenvalue is a root of the characteristic equation will tell us some
information about the matrix when it is compared to another number that will be
discussed later. This number is important enough we give it a name.
Algebraic Multiplicity
If A is a  ×  matrix and   is the characteristic polynomial of the matrix A (i.e.   =
det  −  ) and  is an eigenvalue for A, then  −  is a factor of the polynomial
and the number of times that  −  is a factor of   is called the algebraic multiplicity
of the eigenvalue .
Algebraic multiplicity is the largest value  such that:   = det  −  =  −
Example
For the  ×  matrix to the right find the characteristic
polynomial, all of the eigenvalues for the matrix along with the
algebraic multiplicity of each of the eigenvalues that you found.
The matrix A is diagonal so the characteristic polynomial is:
3
0
0
0
0
0
= 3− 7− 7− 3− 3− 3− = 3−
0
7
0
0
0
0
4
0
0
0
3
0
0
0
0
7
0
0
0
7−
0
0
0
0
3
0
0
0
0
0
0
3
2
The algebraic multiplicity of the eigenvalue 3 is equal to 4.
The eigenvalues for A are : 3,7
The algebraic multiplicity of the eigenvalue 7 is equal to 2.
Example
Find the characteristic polynomial and eigenvalues for the matrix A to
the right.
5
= 0
0
0
6
8
0
0
1
7
1
3
2
4
3
1
Even though A is not triangular take advantage of the zeros in the first two columns.
6
1
2
5−
8−
7
4
8
−

7
4
0
=
= 5−
0
1−
3
0
1−
3
0
0
3
1−
0
3
1−
0
1−
3
= 5− 8−
= 5− 8− 1− 1− −9
3
1−
= 5 −  8 −   2 − 2 − 8
= 5− 8− −4 +2
1st column
expansion
2nd column
expansion
factor
Characteristic polynomial   = 5 −  8 −   − 4  + 2 , Eigenvalues: −2, 4, 5, 8
Properties of the Characteristic Polynomial
We will now look at a few of the more important properties of the characteristic polynomial
of a matrix A.
Characteristic Polynomials and Determinants
The value of the characteristic polynomial at zero is
the determinant of the matrix.
If A is  ×  matrix and
= det  −  then:
0 = det  =
Evaluating Matrix Functions
If you want to plug a  ×  matrix A into a polynomial   (i.e. evaluate   ) this is done
by replacing all constant terms  by  in the polynomial where  is the  ×  identity
matrix. The rest of the polynomial is evaluated using ordinary matrix arithmetic. Being able to
evaluate matrices in polynomials allows fro matrix functions trigonometric, exponential and
logarithmic functions.
Example
For the matrix A to the right find the characteristic polynomial
then evaluate   .
2−
=
3
1
=
−2 −
2 −  −2 −  − 3 =
2 1 2
=  − 7 =
−7
3 −2
7 0
7
=
−
0 7
7
2
2
−4−3=
=
2
−7
1 0
2 1 2 1
7 0
=
−
0 1
3 −2 3 −2
0 7
0
0 0
=
=
0
0 0
Evaluating a Matrix in its Characteristic Polynomial
Its is always true that if you plug a matrix into its characteristic
polynomial you will get the zero matrix.
2
3
1
−2
Characteristic
Equation
The result is
the zero
matrix !
If   = det  −
then:
=
This result has many applications. One in particular is a more efficient way to compute the
inverse of a matrix than augmenting with the identity and row reducing.
Example
Find the characteristic polynomial of the matrix A to the
right and use this to compute −1 .
1 2 −2
= 0 −1 4
0 0
2
Take advantage of the fact that the matrix A is upper triangular so :
= 1 −  −1 −  2 −  =  2 − 1 2 −  = − 3 + 2 2 +  − 2
=
−3 + 22 +  − 2 =
−3 + 22 +  = 2
−2 + 2 +  = 2
1
−2 + 2 +  =
2
−122 +  + 12 =
−1
−1
=
=
1
−2
1
−2
1
0
0
1. The idea here is to move the constant times identity
matrix to one side of the equation.
2. Factor the matrix A out from the other side.
3. Divide by the constant if not 1.
4. Collect all terms in factor other than A.
1 2 −2 1 2 −2
1 2 −2
1 0 0
1
0 −1 4 0 −1 4 + 0 −1 4 + 2 0 1 0
0 0
2 0 0
2
0 0
2
0 0 1
1
0 0
1 2 −3
2
0 2
1 2 −2
1
1 4 + 0 −1 4 + 0 2 0 = 0 −1 2
1
0 0
0 4
0 0
2
0 0 12
2
If 1 , ⋯ ,  eigenvalues for A
then the eigenvalues for 2 are:
12 , ⋯ , 2
Eigenvalues and the Square of a Matrix
If  is an eigenvalue for the matrix A then 2 is
an eigenvalue of the matrix 2 .
The reason for this is as follows. Let v be the eigenvector corresponding to .
2  =  ∙   =   =   =   =  ∙  = 2
Eigenvalues and a Power of a Matrix
This is true for any positive integer p, if  is an eigenvalue
for the matrix A then  is an eigenvalue of the matrix  .
If 1 , ⋯ ,  eigenvalues for A
then the eigenvalues for  are:

1 , ⋯ ,
The reason for this is as follows. Let v be the eigenvector corresponding to .
=  ∙ ⋯ ∙   =  ∙ ⋯ ∙   =  ∙ ⋯ ∙   =   ∙ ⋯ ∙   = ⋯ =

−1
−1
Eigenvalues and a Function of a Matrix
If  is an eigenvalue for the matrix A and   is
any polynomial then   is an eigenvalue of the
matrix   .
−1
If 1 , ⋯ ,  eigenvalues for A and
is a polynomial then the
eigenvalues for   are:
1 , ⋯ ,
The reason for this is as follows. Let v be the eigenvector corresponding to . Let the
polynomial   =    + ⋯ + 1  + 0 .
=   + ⋯ + 1  + 0   =    + ⋯ + 1  + 0
=    + ⋯ + 1  + 0  =   + ⋯ + 1  + 0  =
Example
A 3 × 3 matrix A has eigenvalues of:
− 1, 1, 3. Find the eigenvalues of the
matrices 2 , 3 and 2 − 4 + 3.
Eigenvalues of 2 − 4 + 3 are: −1
The eigenvalues of 2 are: −1 2 , 12 , 32 = 1,9
The eigenvalues of 3 are: −1 3 , 13 , 33 = −1,1,27
2
− 4 −1 + 3, 12 − 4 ∙ 1 + 3, 32 − 4 ∙ 3 + 3 = 0,7
Eigenvalues and the Inverse of a Matrix
If  is an eigenvalue for the matrix A then 1 is
an eigenvalue of the matrix −1 .
If 1 , ⋯ ,  eigenvalues for A then
the eigenvalues for −1 are:
1
1
,
⋯
,

1

The reason for this is as follows. Let v be the eigenvector corresponding to  ≠ 0.
−1  = −1  = −1  = −1   =  =  divide by : −1  = 1
Eigenvalues and the Transpose of a Matrix
The characteristic equation   of a matrix A
is equal to the characteristic equation   of
the matrix  . This means a matrix and its
transpose will have the same eigenvalues.
If 1 , ⋯ ,  eigenvalues for A then the
eigenvalues for  are:
1 , ⋯ ,
The characteristic equation of  and
are equal.
The reason for this is as follows. Remember det  = det  and   = .
= det  −  = det  −   = det  −   = det  −  =
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