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Unformatted text preview: length elements of W1 which are perpendicular
to each other.
Next, we consider the eigenspace W9 corresponding to the eigenvalue λ2 = 9,
which consists of the solutions to the linear system 5−9
4
0
b1
4
5−9
0 b2 = 0,
0
0
1−9
b3
or −4b1
4b1 +4b2
−4b2 −8b3 = 0,
= 0,
= 0. The coeﬃcient matrix of this linear system is −4 4
0 4 −4 0 .
0
0 −8
34 Applying the elementary row operations to this matrix yields −4 4
0
1 −1 0
1 −1 4 −4 0 → 4 −4 0 → 0 0
0
0 −8
0 0 −8
00 1 −1 0
1 −1 0
→ 0 0 0 → 0 0 1 .
001
000 0
0
−8 Thus the linear system is equivalent to
b1 − b2
b3
0
and we see that We set = 0,
= 0,
= 0, 1
W9 = span 1 .
0 b3 = 1
√
2
1
√
2 . 0
Theory guarantees that the matrix 1
0 √2
−
B = 0 √1
2
10 1
√
2
1
√
2 , 0 whose columns are the eigenvectors b1 , b1 , and b3 , will satisfy 100
B −1 AB = 0 1 0 .
009
Moreover, since the eigenvectors we have chosen are of unit length and perpendicular to each other, the matrix B will be orthogonal.
Exercises:
2.1.1. Find a 2 × 2matrix B such that B is orthogonal and B −1 AB is diagonal,
whe...
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 Winter '14
 Equations

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