A First Course in Linear Algebra, 2017a

7.3. Applications of Spectral Theory 371
We verify this works. P T AP is of the form
⎡
0 − 1 ⎤
√ √2 1 ⎡ ⎤
2
1 0 0
⎡
⎢ 1 0 0 ⎥⎢
0 3 1
2 2 ⎥⎢
⎣ 1
0 √2 √2 1 ⎦⎣
⎦⎣
which is the desired diagonal matrix.
⎡
= ⎣
0 1 2
3
2
1 0 0
0 1 0
0 0 2
⎤
⎦
⎤
0 1 0
− √ 1 1
0 √2 ⎥
2 ⎦
1√ 1
2
0 √2
We can now apply this technique to efficiently compute high powers of a symmetric matrix.
♠
Example 7.30: Powers of a Symmetric Matrix
⎡ ⎤
1 0 0
Let A = ⎢ 0 3 1
2 2 ⎥
⎣ ⎦ . Compute A7 .
0 1 2
3
2
Solution. We found in Example 7.29 that P T AP = D is diagonal, where
P =
⎡
⎢
⎣
⎤ ⎡
0 1 0
− √ 1 1
0 √2 ⎥
2 ⎦ and D = ⎣
1√ 1
0 √2
2
1 0 0
0 1 0
0 0 2
Thus A = PDP T and A 7 = PDP T PDP T ···PDP T = PD 7 P T which gives:
⎡
⎡
⎤⎡
⎤
0 1 0
7 0 − 1 ⎤
√ √2 1
1 0 0
2 A 7 ⎢
= ⎣ − √ 1 1
2
0 √2 ⎥
⎦⎣
0 1 0 ⎦
⎢ 1 0 0 ⎥
⎣
1√
2
0
1√ 0 0 2
2
0
1√ √2 1 ⎦
2
⎡
⎡
⎤⎡
⎤ 0 −
0 1 0
1 ⎤
√ √2 1
1 0 0
2 ⎢
= ⎣ − √ 1 2
0
1√ ⎥
2 ⎦⎣
0 1 0 ⎦
⎢ 1 0 0 ⎥
1√ 1
0 √2 0 0 2 7 ⎣ 1
0 √2 √2 1 ⎦
2
=
⎡
⎢
⎣
0 1 0
− √ 1 2
0
1√
2
0
1
⎤
√1
⎥
2 ⎦
√2
⎤
⎦
⎡
0 − √ 1 √2 1 2 ⎢ 1 0 0
⎣
0
2 7 √
2
2 7
⎡
1 0 0
0 27 +1
= ⎢
2
⎣
0 27 −1
2
⎤
⎥
⎦
√
2
2 7 −1
2
2 7 +1
2
⎤
⎥
⎦