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eigenvectors for A even when the roots are repeated.
We ﬁrst consider the eigenspace W1 corresponding to the eigenvalue λ1 = 1,
which consists of the solutions to the linear system 5−1
4
0
b1
4
5−1
0 b2 = 0,
0
0
1−1
b3
or 4b1
4b 1 +4b2
+4b2
0 = 0,
= 0,
= 0. 2 There are many excellent linear algebra texts that prove this theorem in detail; one good
reference is Bill Jacob, Linear algebra , W. H. Freeman, New York, 1990; see Chapter 5. 33 The coeﬃcient matrix of this linear system is 440 4 4 0 .
000
Applying the elementary row operations 440
11 4 4 0 → 4 4
000
00 to this matrix 0
1
0 → 0
0
0 yields
1
0
0 0
0 .
0 Thus the linear system is equivalent to
b1 + b2
0
0 = 0,
= 0,
= 0. Thus W1 is a plane with equation b1 + b2 = 0.
We need to extract two unit length eigenvectors from W1 which are perpendicular to each other. Note that since the equation for W1 is b1 + b2 = 0, the
unit length vector
√ 1
n= 2
1
√
2 0
is perpendicular to W1 . Since √ 1
0
2
−1 b1 = 0 ∈ W1 , we ﬁnd that b2 = n × b1 = √2
∈ W1 .
1
0
The vectors b1 and b2 are unit...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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