Multiplicity and Diagonalization 225

A Diagonalizable Matrix with an Eigenvalue Having Multiplicity 2

? Problem

Use the following matrix in both problems:

1

1

−2

0

1

0

0

−1

3

1. Find all eigenvalues and eigenvectors of the matrix.

2. Express the matrix in the following form:

x

11

x

21

x

31

x

12

x

22

x

32

x

13

x

23

x

33

x

11

x

21

x

31

x

12

x

22

x

32

x

13

x

23

x

33

λ

1

0

0

0

λ

2

0

0

0

λ

3

−1

! Solution

1. The eigenvalues

λ

of the 3×3 matrix

1

1

−2

0

1

0

0

−1

3

are the roots of the characteristic equation:

1 − λ

1

−2

0

1 − λ

0

0

−1

3 − λ

det = 0.

λ = 3, 1

= (1 − λ)(1 − λ)(3 − λ) + 0 · (−1) · (−2) + 0 · 1 · 0

− 0 · (1 − λ) · (−2) − 0 · 1 · (3 − λ) − (1 − λ) · (−1) · 0

= (1 − λ)

2

(3 − λ) = 0

1 − λ

1

−2

0

1 − λ

0

0

−1

3 − λ

det

Note that the eigenvalue 1 has multiplicity 2.

226 Chapter 8 Eigenvalues and Eigenvectors

a.

The eigenvectors corresponding to

λ

= 3

Let’s insert our eigenvalue into the following formula:

= λ

=

0

0

0

1 − λ

1

−2

0

1 − λ

0

0

−1

3 − λ

x

1

x

2

x

3

x

1

x

2

x

3

x

1

x

2

x

3

1

1

−2

0

1

0

0

−1

3

, that is

This gives us:

1 − 3

1

−2

0

1 − 3

0

0

−1

3 − 3

−2

1

−2

0

−2

0

0

−1

0

−2x

1

x

1

− 2x

2

− x

3

−2x

1

= = =

x

1

x

2

x

3

x

1

x

2

x

3

0

0

0

The solutions are as follows:

x

1

= 0

x

3

= −2x

2

=

= c

1

0

c

1

−2c

1

0

1

−2

x

1

x

2

x

3

and the eigenvector

where c

1

is a real nonzero number.

b. The eigenvectors corresponding to

λ

= 1

Repeating the steps above, we get

1 − 1

1

−2

0

1 − 1

0

0

−1

3 − 1

0

1

−2

0

0

0

0

−1

2

0

x

1

− x

3

−2x

1

+ 2x

3

= = =

x

1

x

2

x

3

x

1

x

2

x

3

0

0

0

and see that x

3

= x

1

and x

2

can be any real number. The eigenvector conse-

quently becomes

=

= c

1

+ c

2

x

1

x

2

x

3

c

1

c

2

c

1

1

0

1

0

1

0

where c

1

and c

2

are arbitrary real numbers that cannot both be zero.

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