A ¼
A
nÀ1
ð
Þ
a
a
T
a n
!
,
where A
(n À 1) is a symmetric matrix and non-singular as well. We define P such that
P ¼
E A
nÀ1
ð
Þ
À1 a
0
1
!
,
where E is a (n À 1, n À 1) unit matrix. Notice that detP ¼ det E Á 1 ¼ 1 6 ¼ 0,
indicating that P is non-singular. We have
P
T
¼
E
0
a
T A
nÀ1
ð
Þ
À1
1
:
For this expression, consider a non-singular matrix S. Then, we have SS
À1
¼ E.
Taking its transposition, we have (S
À1 )
T S
T
¼ E. Therefore, if S is a symmetric matrix
(S
À1 )
T S ¼ E; i.e., (S
À1 )
T
¼ S
À1 . Hence, an inverse matrix of a symmetric matrix is
symmetric as well. Then, for a symmetric matrix A
(n À 1) we have
a
T A
nÀ1
ð
Þ À1
T ¼ A
nÀ1
ð
Þ À1
T a
T
À Á T ¼ A
nÀ1
ð
Þ À1 a:
Therefore, A can be expressed as
A ¼ P
T
A
nÀ1
ð
Þ
0
0
a n À A
nÀ1
ð
Þ
À1 a
½ Š
!
P
¼
E
0
a
T A
nÀ1
ð
Þ
À1
1
!
A
nÀ1
ð
Þ
0
0
a n À A
nÀ1
ð
Þ
À1 a
½ Š
!
E A
nÀ1
ð
Þ
À1 a
0
1
!
:
ð14:100Þ
Now, taking a determinant of (14.100), we have
det A ¼ det P
T det A
nÀ1
ð
Þ
h
i
a n À A
nÀ1
ð
Þ À1 a
½ Š
n
o
det P
¼ det A
nÀ1
ð
Þ
h
i
a n À A
nÀ1
ð
Þ À1 a
½ Š
n
o
:
By supposition, we have det A
(n À 1)
> 0 and det A > 0. Hence, we have
14.5 Hermitian Quadratic Forms
571