1.3 Solutions
47
1.31 A =
⎛
⎝
6 −2 2
−2 3 −1
2 −1 3
⎞
⎠
In Problem 1.30, the characteristic roots are found to be λ = 2, 2, 8. With
λ = 2, we find the invariant vectors.
⎛
⎝
6 − 2 −2
2
−2 3 − 2 −1
2 −1 3 − 2
⎞
⎠
⎛
⎝
x 1
x 2
x 3
⎞
⎠ = 0
The two vectors are X 1 = (1, 1, −1)
and X 2 = (0, 1, 1)
. The third vector
can be obtained in a similar fashion. It can be chosen as X 3 = (2, −1, 1)
. The
three column vectors can be normalized and arranged in the form of a matrix.
The matrix A is diagnalized by the similarity transformation.
S
−1 AS = diag A
S =
⎛
⎜
⎝
1
√
3
0
2
√
6
1
√
3
1
√
2
−
1
√
6
−
1
√
3
1
√
2
1
√
6
⎞
⎟
⎠
As the matrix S is orthogonal, S
−1
= S
. Thus
⎛
⎜
⎝
1
√
3
1
√
3
−
1
√
3
0
1
√
2
1
√
2
2
√
6
−
1
√
6
1
√
6
⎞
⎟
⎠
⎛
⎝
6 −2 2
−2 3 −1
2 −1 3
⎞
⎠
⎛
⎜
⎝
1
√
3
0
2
√
6
1
√
3
1
√
2
−
1
√
6
−
1
√
3
1
√
2
1
√
6
⎞
⎟
⎠ =
⎛
⎝
2 0 0
0 2 0
0 0 8
⎞
⎠
1.32 H =
a 11 a 12
a 21 a 22
A =
3 2
4 1
(a)
3 − λ 2
4 1 − λ
= 0, characteristic equation is
(3 − λ)(1 − λ) − 8 = 0
λ
2
− 4λ − 5 = 0, (λ − 5)(λ + 1) = 0
The eigen values are λ 1 = 5 and λ 2 = −1
(b) and (c) The desired matrix has the form
C =
C 11 C 12
C 21 C 22
The columns which satisfy the system of equations
(a i j − δ i j λ k )C jk = 0, no sum on k
(1)
yielding
(a 11 − λ k )C 1k + a 12 C 2k = 0, no sum on k
a 21 C 1k + (a 22 − λ k )C 2k = 0, k = 1, 2
Since a 11 = 3, a 21 = 4, a 12 = 2, a 22 = 1, we get