Linear Algebra, Theory And Applications, 2012a

13.11. THE MOORE PENROSE INVERSE 333
Let
(
V ∗ P Q
A 0 U =
R S
where P is k × k.
Next use the first equation of (13.25) to write
)
(13.26)
{ ( }} ) { A
P Q
V
U ∗
R S
A 0
A
{ }} {
UΣV ∗
{ }} {
UΣV ∗ =
A
{ }} {
UΣV ∗ .
Then multiplying both sides on the left by U ∗ and on the right by V,
( )( )( ) ( )
σ 0 P Q σ 0 σ 0
=
0 0 R S 0 0 0 0
Now this requires ( σPσ 0
0 0
)
=
( σ 0
0 0
Therefore, P = σ −1 . From the requirement that AA 0 is Hermitian,
)
. (13.27)
{ ( }} {
P Q
V
R S
A 0
A
{ }} {
UΣV ∗
)
U ∗ = U
( σ 0
0 0
)( P Q
R S
)
U ∗
must be Hermitian. Therefore, it is necessary that
( )( )
σ 0 P Q
=
0 0 R S
=
( )
σP σQ
0 0
( ) I σQ
0 0
is Hermitian. Then ( I σQ
0 0
)
=
( I 0
Q ∗ σ 0
Thus
Q ∗ σ =0
and so multiplying both sides on the right by σ −1 , it follows Q ∗ = 0 and so Q =0.
From the requirement that A 0 A is Hermitian, it is necessary that
)
A 0
{ ( }} ) { A
P Q
V
U ∗
R S
{ }} {
UΣV ∗ = V
= V
( )
Pσ 0
Rσ 0
(
I 0
Rσ 0
V ∗
)
V ∗
is Hermitian. Therefore, also (
I 0
Rσ 0
is Hermitian. Thus R = 0 because this equals
( ) ∗ ( I 0 I σ
=
∗ R ∗
Rσ 0 0 0
)
)