Principal minors of H are 5 (>0) and 1 (>0) and detH ¼ 5 À 4 ¼ 1 (>0).
Therefore, from Theorem 14.11 we have H > 0. A characteristic equation gives
following eigenvalues; i.e.,
λ 1 ¼ 3 þ 2
ffiffi ffi
2
p
, λ 2 ¼ 3 À 2
ffiffi ffi
2
p :
Both the eigenvalues are positive as anticipated. As a diagonalizing matrix R, we
get
R ¼
1 þ
ffiffi ffi
2
p
1 À
ffiffi ffi
2
p
1
1
!
:
To obtain a unitary matrix, we have to seek norms of column vectors.
Corresponding to λ 1 and λ 2 , we estimate their norms to be
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 þ 2
ffiffi ffi
2
p
p
and
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 À 2
ffiffi ffi
2
p
p
, respectively. Using them, as a unitary matrix U we get
U ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 À 2
ffiffi ffi
2
p
p
À
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 þ 2
ffiffi ffi
2
p
p
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 þ 2
ffiffi ffi
2
p
p
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 À 2
ffiffi ffi
2
p
p
0
B
B
B
@
1
C
C
C
A
:
ð14:102Þ
Thus, performing the matrix diagonalization, we obtain a diagonal matrix D such
that
D ¼ U
{ HU ¼
3 þ 2
ffiffi ffi
2
p
0
0
3À 2
ffiffi ffi
2
p
!
:
ð14:103Þ
Let us view the above unitary diagonalization in terms of coordinate transformation. Using the above matrix U and changing (14.101) as in (13.36),
x H
j jx
h
i¼ x 1 x 2
ð
ÞUU
{ 5 2
2 1
UU
{ x 1
x 2
¼ x 1 x 2
ð
ÞU
3 þ 2
ffiffi ffi
2
p
0
0
3À 2
ffiffi ffi
2
p
!
U
{ x 1
x 2
:
ð14:104Þ
Making an argument analogous to that of Sect. 13.2 and using similar notation,
we have
14.5 Hermitian Quadratic Forms
573