1 À λλ
Ã
ð
Þjλi ¼ 0:
ð14:90Þ
As jλi 6 ¼ 0 is assumed, 1 À λλ
à ¼ 0. That is
λλ
Ã
¼ λ
j j
2 ¼ 1:
ð14:91Þ
Thus, eigenvalues of a unitary matrix have unit absolute value. Let those eigenvalues be λ k ¼ e
iθ k k ¼ 1, 2, Á Á Á; θ k : real
ð
Þ . Then, from (12.11) we have
detU ¼
Y
k
λ k ¼ e
i θ 1 þθ 2 þÁÁÁ
ð
Þ and detU
j
j¼
Y
k
λ k
j j ¼
Y
k
e
iθ k
¼ 1:
That is, any unitary matrix has a determinant of unit absolute value.
∎
Example 14.2 Let us think of a following unitary matrix R:
R ¼
cos θ À sin θ
sin θ
cos θ
:
ð14:92Þ
A characteristic equation is
cos θ À λ
À sin θ
sin θ
cos θ À λ
¼ λ
2
À 2λ cos θ þ 1:
ð14:93Þ
Solving (14.93), we have
λ ¼ cos θ Æ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
cos 2 θ À 1
p
¼ cos θ Æ i j sin θ j :
ð14:94Þ
(i) θ ¼ 0: This is a trivial case. The matrix R has automatically been diagonalized
to be an identity matrix. Eigenvalues are 1 (double root).
(ii) θ ¼ π: This is a trivial case. Eigenvalues are À1 (double root).
(iii) θ 6 ¼ 0, π: Let us think of 0 < θ < π. Then λ ¼ cos θ Æ i sin θ. As a diagonalizing
unitary matrix U, we get
U ¼
1
ffiffi ffi
2
p
1
ffiffi ffi
2
p
À
i ffiffi ffi
2
p
i ffiffi ffi
2
p
0
B
B
@
1
C
C
A , U
{
¼
1
ffiffi ffi
2
p
i ffiffi ffi
2
p
1
ffiffi ffi
2
p
À
i ffiffi ffi
2
p
0
B
B
@
1
C
C
A :
ð14:95Þ
Therefore, we have
14.4 Hermitian Matrices and Unitary Matrices
567