A First Course in Linear Algebra, 2017a

577
7.4.12 The eigenvectors and eigenvalues are:
⎧⎡
⎨ − 1 ⎤ ⎡ √
6√
6
1
⎤⎫
⎧⎡
3√
2 3 ⎬ ⎨
⎣ 0
⎩ √ √
⎦, ⎣ 1
√5√ √ 5 ⎦
⎭ ↔−1, ⎣
⎩
5 6 2 15
The columns of U are these vectors.
⎡
⎣
1
6
1
15
− 1 √ √ √
6√
6
1
3
2 3
1
⎤
√ 6 √ 6
1
0 5 5 −
2
√ √ 5 5 ⎦
√ √ √
5 6
1
15
2 15
1
30
30
1
6
T
1
⎤⎫
6√
6 ⎬
− 2 5√
√ 5 ⎦
⎭ ↔ 2.
30
1
30
⎡
− 1 2
− 1 √ √ √
5 6 5
1
10
5
− 1 √
5√
6 5
7
5
− 1 5√
6
⎢
⎣ √ √
5 −
1
5
6 −
9
1
10
⎤
⎥
⎦
10
·
⎡
⎣
− 1 √ √ √
6√
6
1
3
2 3
1
⎤ ⎡
√ 6 √ 6
1
0 5 5 −
2
√ √ 5 5 ⎦
√ √ √
= ⎣
5 6
1
15
2 15
1
30
30
1
6
−1 0 0
0 −1 0
0 0 2
⎤
⎦
7.4.13 If A is given by the formula, then
A T = U T D T U = U T DU = A
Next suppose A = A T . Then by the theorems on symmetric matrices, there exists an orthogonal matrix U
such that
UAU T = D
for D diagonal. Hence
7.4.14 Since λ ≠ μ, itfollowsX •Y = 0.
A = U T DU
7.4.21 Using the QR factorization, we have:
⎡ ⎤ ⎡ √ √ √
1
1 2 1
6 6
3
10
2
7
⎤⎡
√ √ 15 √ 3
⎣ 2 −1 0 ⎦ = ⎣ 1
3 6 −
2
5
2 −
1
15
3 ⎦⎣
√ √ √
1 3 0 6
1
2
2 −
1
3
3
A solution is then
1
1
1
6
1
6
⎡ ⎤ ⎡
6√
6
3
⎤ ⎡
10√
2
⎣
3√
√ 6 ⎦, ⎣ − 2 5√
√ 2 ⎦, ⎣
6 2
1
2
7
15√
3
−
15√ 1 3
− 1 3
√ √ √
6
1
2
6
1
⎤
√ 6 √ 6
5
0 2 2
3
10
2 ⎦
√
7
0 0 15 3
⎤
⎦
√
3
7.4.22 ⎡
⎢
⎣
1 2 1
2 −1 0
1 3 0
0 1 1
⎤ ⎡
⎥
⎦ = ⎢
⎣
√ √ √ √ √ √
6
1
6
2 3
5
111 3 37
7
⎤
√ √ √ √ √ 111
6 −
2
9
2 3
1
333 3 37 −
2
√ 111√
111
√ √ √ √ √ ⎥
6
5
18
2 3 −
17
333 3 37 −
1
√ 37 111 ⎦
√ √ √ √
1
0 9 2 3
22
333 3 37 −
7
111 111
1
6
1
3
1
6