1.3 Solutions 47
1.31 A=
⎛
⎝
6 − 22
− 23 − 1
2 − 13
⎞
⎠
In Problem 1.30, the characteristic roots are found to beλ= 2 , 2 ,8. With
λ=2, we find the invariant vectors.
⎛
⎝
6 − 2 − 22
− 23 − 2 − 1
2 − 13 − 2
⎞
⎠
⎛
⎝
x 1
x 2
x 3
⎞
⎠= 0
The two vectors areX 1 =(1, 1 ,−1)′andX 2 =(0, 1 ,1)′. The third vector
can be obtained in a similar fashion. It can be chosen asX 3 =(2,− 1 ,1)′.The
three column vectors can be normalized and arranged in the form of a matrix.
The matrixAis diagnalized by the similarity transformation.
S−^1 AS=diagA
S=
⎛
⎜
⎝
√^1
3 0
√^2
6
√^1
3
√^1
2 −
√^1
6
−√^13 √^12 √^16
⎞
⎟
⎠
As the matrixSis orthogonal,S−^1 =S′. Thus
⎛
⎜
⎝
√^1
3
√^1
3 −
√^1
3
0 √^12 √^12
√^2
6 −
√^1
6
√^1
6
⎞
⎟
⎠
⎛
⎝
6 − 22
− 23 − 1
2 − 13
⎞
⎠
⎛
⎜
⎝
√^1
3 0
√^2
6
√^1
3
√^1
2 −
√^1
6
−√^13 √^12 √^16
⎞
⎟
⎠=
⎛
⎝
200
020
008
⎞
⎠
1.32 H=
(
a 11 a 12
a 21 a 22
)
A=
[
32
41
]
(a)
∣
∣
∣
∣
3 −λ 2
41 −λ
∣
∣
∣
∣=0, characteristic equation is
(3−λ)(1−λ)− 8 = 0
λ^2 − 4 λ− 5 = 0 ,(λ−5)(λ+1)= 0
The eigen values areλ 1 =5 andλ 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
(aij−δijλk)Cjk= 0 ,no sum onk (1)
yielding
(a 11 −λk)C 1 k+a 12 C 2 k=^0 ,no sum onk
a 21 C 1 k+(a 22 −λk)C2k= 0 ,k= 1 , 2
Sincea 11 =3,a 21 =4,a 12 =2,a 22 =1, we get
