Linear Algebra. Chapter 5. Eigenvalues and Eigenvectors

1. Chapter 5 Eigenvalues and Eigenvectors

Linear Algebra
Chapter 5
Eigenvalues and Eigenvectors

2. Eigenvalues and Eigenvectors

Definition
Let A be an n n matrix. A scalar is called an eigenvalue of A
if there exists a nonzero vector x in Rn such that
Ax = x.
The vector x is called an eigenvector corresponding to .
Figure 5.1
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3. Computation of Eigenvalues and Eigenvectors

Let A be an n n matrix with eigenvalue and corresponding
eigenvector x. Thus Ax = x. This equation may be written
Ax – x = 0
given
(A – In)x = 0
Solving the equation |A – In| = 0 for leads to all the eigenvalues
of A.
On expanding the determinant |A – In|, we get a polynomial in .
This polynomial is called the characteristic polynomial of A.
The equation |A – In| = 0 is called the characteristic equation of
A.
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4. Example 1

Find the eigenvalues and eigenvectors of the matrix
4 6
A
5
3
Solution Let us first derive the characteristic polynomial of A.
We get
4 6
1 0 4 6
A I 2
3
5
0
1
3
5
A I 2 ( 4 )(5 ) 18 2 2
We now solve the characteristic equation of A.
2 2 0 ( 2)( 1) 0 2 or 1
The eigenvalues of A are 2 and –1.
The corresponding eigenvectors are found by using these values
of in the equation(A – I2)x = 0. There are many eigenvectors
corresponding to each eigenvalue.
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5.

For = 2
We solve the equation (A – 2I2)x = 0 for x.
The matrix (A – 2I2) is obtained by subtracting 2 from the
diagonal elements of A. We get
6 6 x1
0
3
3 x2
This leads to the system of equations
6 x1 6 x2 0
3x1 3x2 0
giving x1 = –x2. The solutions to this system of equations are
x1 = –r, x2 = r, where r is a scalar. Thus, the eigenvectors of A
corresponding to = 2 are nonzero vectors of the form
x1
1
1
v1 x2 r
1
1
x2
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6.

For = –1
We solve the equation (A + 1I2)x = 0 for x.
The matrix (A + 1I2) is obtained by adding 1 to the diagonal
elements of A. We get
3 6 x1
0
3
6 x2
This leads to the system of equations
3x1 6 x2 0
3x1 6 x2 0
Thus x1 = –2x2. The solutions to this system of equations are x1
= –2s and x2 = s, where s is a scalar. Thus, the eigenvectors of
A corresponding to = –1 are nonzero vectors of the form
x1
2 2
v2 x2 s
1 1
x2
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7. Theorem

Let A be an n n matrix and an eigenvalue of A. The set of all
eigenvectors corresponding to , together with the zero vector, is
a subspace of Rn. This subspace is called the eigenspace of .
Proof
Let x1 and x2 be two vectors in the eigenspace of and let c be a
scalar. Then Ax1 = x1 and Ax2 = x2. Hence,
Ax1 Ax 2 x1 x 2
A(x1 x 2 ) (x1 x 2 )
Thus x1 x 2 is a vector in the eigenspace of . The set is closed
under addition.
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8.

Further, since Ax1 = x1,
cAx1 c x1
A(cx1 ) (cx1 )
Therefore, cx1 is a vector in the eigenspace of . The set is
closed scalar multiplication.
Thus, the set is a subspace of Rn.
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9. Example 2

Find the eigenvalues and eigenvectors of the matrix
5 4 2
A 4 5 2
2 2 2
Solution The matrix A – I3 is obtained by subtracting from
the diagonal elements of A.Thus
2
5 4
A I 3 4 5 2
2
2
2
The characteristic polynomial of A is |A – I3|. Using row and
column operations to simplify determinants, we get
5
4
A I 3 4 5
2
2
2
1
2 4
2
2
1
5
2
0
2
2
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10.

1
0
4 9
2
4
0
2
2
(1 )[(9 )( 2 ) 8] (1 )[ 2 11 10]
(1 )( 10)( 1) ( 10)( 1)
2
We now solving the characteristic equation of A:
( 10)( 1) 2 0
10 or 1
The eigenvalues of A are 10 and 1.
The corresponding eigenvectors are found by using three values
of in the equation (A – I3)x = 0.
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11.

1 = 10
We get
( A 10 I 3 )x 0
5 4 2 x1
4 5 2 x2 0
2 2 8 x3
The solution to this system of equations are x1 = 2r, x2 = 2r,
and x3 = r, where r is a scalar.
Thus the eigenspace of 1 = 10 is the one-dimensional space
of vectors of the form.
2
r 2
1
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12.

2 = 1
Let = 1 in (A – I3)x = 0. We get
( A 1I 3 )x 0
4 4 2 x1
4 4 2 x2 0
2 2 1 x3
The solution to this system of equations can be shown to be
x1 = – s – t, x2 = s, and x3 = 2t, where s and t are scalars.
Thus the eigenspace of 2 = 1 is the space of vectors of the
form.
s t
s
2t
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13.

Separating the parameters s and t, we can write
s t 1 1
s s 1 t 0
2t 0 2
Thus the eigenspace of = 1 is a two-dimensional subspace of
R3 with basis
1 1
1 , 0
0 0
If an eigenvalue occurs as a k times repeated root of the
characteristic equation, we say that it is of multiplicity k.
Thus =10 has multiplicity 1, while =1 has multiplicity 2
in this example.
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14. Homework

Exercises will be assigned by the teachers of
the practice classes.
Ex: Prove that if A is a diagonal matrix, then its eigenvalues are
the diagonal elements.
Ex: Prove that if A and At have the same eigenvalues.
Ex: Prove that the constant term of the characteristic polynomial
of a matrix A is |A|.
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15. Diagonalization of Matrices

Definition
Let A and B be square matrices of the same size. B is said to be
similar to A if there exists an invertible matrix C such that
B = C–1AC. The transformation of the matrix A into the matrix B
in this manner is called a similarity transformation.
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16. Example 3

Consider the following matrices A and C with C is invertible.
Use the similarity transformation C–1AC to transform A into a
matrix B.
7 10
2 5
A
C
3 4
1 3
Solution
1
2 5 7 10 2 5
1
B C AC
1 3 3 4 1 3
3 5 7 10 2 5
1 2 3 4 1 3
6 10 2 5
1 2 1 3
2 0
0 1
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17. Theorem

Similar matrices have the same eigenvalues.
Proof
Let A and B be similar matrices. Hence there exists a matrix C
such that B = C–1AC.
The characteristic polynomial of B is |B – In|. Substituting for B
and using the multiplicative properties of determinants, we get
B I C 1 AC I C 1 ( A I )C
C 1 A I C A I C 1 C
A I C 1C A I I
A I
The characteristic polynomials of A and B are identical. This
means that their eigenvalues are the same.
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18.

Definition
A square matrix A is said to be diagonalizable if there exists a
matrix C such that D = C–1AC is a diagonal matrix.
Theorem 5.4
Let A be an n n matrix.
(a) If A has n linearly independent eigenvectors, it is
diagonalizable. The matrix C whose columns consist of n
linearly independent eigenvectors can be used in a similarity
transformation C–1AC to give a diagonal matrix D. The
diagonal elements of D will be the eigenvalues of A.
(b) If A is diagonalizable, then it has n linearly independent
eigenvectors
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19. Example 4

4 6
(a) Show that the matrix A 3 5 is diagonalizable.
(b) Find a diagonal matrix D that is similar to A.
(c) Determine the similarity transformation that diagonalizes A.
Solution
(a) The eigenvalues and corresponding eigenvector of this
matrix were found in Example 1 of Section 5.1. They are
1
1 2, v1 r
1
2
2 1, and v 2 s
1
Since A, a 2 2 matrix, has two linearly independent
eigenvectors, it is diagonalizable.
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20.

(b) A is similar to the diagonal matrix D, which has diagonal
elements 1 = 2 and 2 = –1. Thus
4 6
2 0
A
is similar to D
3
5
0
1
(c) Select two convenient linearly independent eigenvectors, say
1
2
v1 and v 2
1
1
Let these vectors be the column vectors of the diagonalizing
matrix C.
1 2
C
1
1
We get
1
1 2 4 6 1 2
1
C AC
3
1
1
2
5
1
1 2 4 6 1 2 2 0
D
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5 1
1 0 1
1 1 3

21.

Note
If A is similar to a diagonal matrix D under the transformation
C–1AC, then it can be shown that Ak = CDkC–1.
This result can be used to compute Ak. Let us derive this result
and then apply it.
This leads to
Ak CD k C 1
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22. Example 5

Compute A9 for the following matrix A.
4 6
A
5
3
Solution
A is the matrix of the previous example. Use the values of C and
D from that example. We get
9
0 512 0
2 0 29
9
D
0 ( 1)9 0 1
0
1
A9 CD 9C 1
1
1 2 512 0 1 2
514 1026
1
513
1 0 1 1
1
1025
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23. Example 6

Show that the following matrix A is not diagonalizable.
5 3
A
3 1
Solution
5
3
A I2
3
1
The characteristic equation is
A I 2 0 (5 )( 1 ) 9 0
2 4 4 0 ( 2)( 2) 0
There is a single eigenvalue, = 2. We find he corresponding
eigenvectors. (A – 2I ) x = 0 gives 3 3 x1
3 3 x 0 3x1 3x2 0.
2
Thus x1 = r, x2 = r. The eigenvectors are nonzero vectors of the
1
form
r
1
The eigenspace is a one-dimensional space. A is a 2 2
matrix, but it does not have two linearly independent
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eigenvectors. Thus A is not diagonalizable.

24. Spectral Theorem

Let A be an n n symmetric matrix.
(a) All the eigenvalues of A are real numbers.
(b) The dimension of an eigenspace of A is the multiplicity of the
eigenvalues as a root of the characteristic equation.
(c) A has n linearly independent eigenvectors.
Exercises will be assigned by the teachers of the
practice classes.
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