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The Manga Guide to Linear Algebra by Ltd. Trend-Pro Co., Iroha Inoue, Shin Takahashi

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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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