1.3 Solutions
45
H X i = λ i X i
(1)
Now ¯
X
ι H X i = ¯
X
ι λ i X i = λ i ¯
X
ι X i
(2)
is real and non-zero. Similarly the conjugate transpose
¯
X
ι H X i = ¯
λ ι ¯
X
ι X i
(3)
Comparing (2) and (3),
¯
λ i = λ i
Thus λ i is real
1.28 The characteristic equation is given by
1 − λ −1
1
0 3 − λ −1
0
0 2− λ
= 0
(1 − λ)(3 − λ)(2 − λ) + 0 + 0 = 0
( 1 )
or λ
3
− 6λ
2
+ 11λ − 6 = 0 (characteristic equation)
(2)
The eigen values are λ 1 = 1, λ 2 = 3, and λ 3 = 2.
1.29 Let X =
x
y
AX =
−1 0
0 1
x
y
=
−x
−y
It produces reflection through the origin, that is inversion. A performs the
parity operation, Fig. 1.10a.
Fig. 1.10a Parity operation
(inversion through origin)
B X =
0 1
1 0
x
y
=
y
x
Here the x and y coordinates are interchanged. This is equivalent to a reflection about a line passing through origin at θ = 45
◦ , Fig. 1.10b
C X =
2 0
0 2
x
y
=
2x
2y
45
H X i = λ i X i
(1)
Now ¯
X
ι H X i = ¯
X
ι λ i X i = λ i ¯
X
ι X i
(2)
is real and non-zero. Similarly the conjugate transpose
¯
X
ι H X i = ¯
λ ι ¯
X
ι X i
(3)
Comparing (2) and (3),
¯
λ i = λ i
Thus λ i is real
1.28 The characteristic equation is given by
1 − λ −1
1
0 3 − λ −1
0
0 2− λ
= 0
(1 − λ)(3 − λ)(2 − λ) + 0 + 0 = 0
( 1 )
or λ
3
− 6λ
2
+ 11λ − 6 = 0 (characteristic equation)
(2)
The eigen values are λ 1 = 1, λ 2 = 3, and λ 3 = 2.
1.29 Let X =
x
y
AX =
−1 0
0 1
x
y
=
−x
−y
It produces reflection through the origin, that is inversion. A performs the
parity operation, Fig. 1.10a.
Fig. 1.10a Parity operation
(inversion through origin)
B X =
0 1
1 0
x
y
=
y
x
Here the x and y coordinates are interchanged. This is equivalent to a reflection about a line passing through origin at θ = 45
◦ , Fig. 1.10b
C X =
2 0
0 2
x
y
=
2x
2y