A First Course in Linear Algebra, 2017a

557
4.11.6 (a) Orthogonal.
(b) Symmetric.
(c) Skew Symmetric.
4.11.7 ‖U⃗x‖ 2 = U⃗x •U⃗x = U T U⃗x •⃗x = I⃗x •⃗x = ‖⃗x‖ 2 .
Next suppose distance is preserved by U.Then
(U (⃗x +⃗y)) • (U (⃗x +⃗y)) = ‖Ux‖ 2 + ‖Uy‖ 2 + 2(Ux•Uy)
= ‖⃗x‖ 2 + ‖⃗y‖ 2 + 2 ( U T U⃗x •⃗y )
But since U preserves distances, it is also the case that
(U (⃗x +⃗y) •U (⃗x +⃗y)) = ‖⃗x‖ 2 + ‖⃗y‖ 2 + 2(⃗x •⃗y)
Hence
and so
⃗x •⃗y = U T U⃗x •⃗y
((
U T U − I ) ⃗x ) •⃗y = 0
Since y is arbitrary, it follows that U T U − I = 0. Thus U is orthogonal.
4.11.8 You could observe that det ( UU T ) =(det(U)) 2 = 1sodet(U) ≠ 0.
4.11.9
=
This requires a = 1/ √ 3,b = 1/ √ 3.
⎡
⎢
⎣
−1 √
2
1√
2
0
−1 √
6
⎡
−1 √
2
−1 √
6
1 √3
⎤⎡
−1 √
2
−1 √
6
1√ √−1
⎢ a
1√ √−1
⎣ 2 6 ⎥⎢
a
√ ⎦⎣
2 6 ⎥
√ ⎦
0 6
3
b 0 6
3
b
⎡
√
1
1 3√
3a −
1 1
3 3
3b −
1
⎤
3
⎣ 1
3√
3a −
1
3
a 2 + 2 3
ab − 1 ⎦
3
1
3√
3b −
1
3
ab − 1 3
b 2 + 2 3
1 √3
⎤⎡
√−1
6
1/ √ 3
√
6
3
1/ √ ⎥⎢
⎦⎣
3
−1 √
2
1√
2
0
−1 √
6
1 √3
⎤
√−1
6
1/ √ 3
√
6
3
1/ √ ⎥
⎦
3
T
1 √3
⎡
= ⎣
⎤
T
1 0 0
0 1 0
0 0 1
⎤
⎦
4.11.10
⎡
⎢
⎣
2
3
2
3
√
2
2
1
6
√
2
− √ 2
2
a
− 1 3
0 b
⎤⎡
⎥⎢
⎦⎣
2
3
2
3
√
2
2
1
6
√
2
− √ 2
2
a
− 1 3
0 b
⎤
⎥
⎦
T
⎡
= ⎣
1
6
√
1
2a −
1
√ √
1
6 2a −
1 1
18 6
2b −
2
9
√ 18
a 2 + 17
18
ab − 2 9
1
6
2b −
2
9
ab − 2 9
b 2 + 1 9
⎤
⎦