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[ 406
]
A GENERALIZED INVERSE FOR MATRICES
BY R. PENROSE
Communicated by
J. A.
TODD
Received
26 July 1954
This paper describes
a generalization of the inverse of a nonsingular matrix, as the
unique solution of
a certain set of equations. This generalized inverse exists for any
(possibly rectangular) matrix whatsoever with complex elements J. It is used here for
solving linear matrix equations, and among other applications for finding an expression
for the principal idempotent elements of a matrix. Also a new type of spectral decom
position is given.
In another paper
its application to substitutional equations and the value of
hermitian idempotents will be discussed.
Notation.
Capital letters always denote matrices (not necessarily square) with com
plex elements. The conjugate transpose of the matrix
A
is written
A*.
Small letters
are used for column vectors (with an asterisk for row vectors) and small Greek letters
for complex numbers.
The following properties of the conjugate transpose will be used:
A** =
A,
(A+B)* = A* + B*,
(BA)*
=
A*B*,
A A*
= 0
implies
^4
= 0.
The last of these follows from the fact that the trace of
A A*
is the sum of the squares
of the moduli of the elements of
A.
Erom the last two we obtain the rule
BAA*
=
CAA* implies BA
=
GA,
(1)
since
(BAA*CAA*)(BC)* = (BACA)(BAGA)*.
Similarly
BA*A
=
CA*A implies BA*
=
GA*.
(2)
THEOREM
\. The four equations
AX
A
—
A
C\\
XAX
=
X,
(4)
{AX)*
=
AX,
(5)
{XA)*
=
XA,
(6)
have a unique solution for any
A.
%
Matrices over more general rings will be considered in
a later paper.
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View Full Document A generalized inverse
for matrices
407
Proof.
I first show that equations (4) and (5) are equivalent to the single equation
XX*A*
=
X.
(7)
Equation
(7) follows from (4) and (5), since it is merely (5) substituted in (4). Con
versely,
(7) implies
AXX*A*
=
AX,
the lefthand side of which is hermitian. Thus
(5) follows, and substituting (5)
in (7) we get (4). Similarly, (3) and (6) can be replaced
by
the equation
XAA*
= A*.
(8)
Thus
it is sufficient to find an
X
satisfying (7)
and (8). Such an
X
will exist
if a
B
can
be found satisfying
M
, ,
3
B
BA*AA*
= A*.
For then
X
=
BA*
satisfies
(8). Also, we have seen that (8) implies
A*X*A*
=
A*
and therefore
BA*X*A*
=
BA*.
Thus
X
also satisfies
(7).
Now
the expressions
A*A, (A*A)
2
, (A*A)
3
,...
cannot
all be linearly independent,
i.e. there exists
a relation
A
1
A*A + \
2
(A*A)*+.
..+A
k
(A*A)
k
= 0
>
(9)
where A,,.
.., A
ft
are not all zero. Let A, be the first nonzero A and put
B
=

VH
Thus
(9) gives
B(A*A)
r+1
= (A*A)
T
,
and applying (1) and (2) repeatedly we obtain
BA*AA*
=
A*,
as required.
To show that
X
is unique, we suppose that
X
satisfies (7)
and (8) and that
Y
satisfies
Y
= A*Y*Y
and
A* = A*AY.
These last relations
are obtained by respectively
substituting (6)
in (4) and (5) in (3). (They are (7) and (8) with
Y
in place of X and the
reverse order
of multiplication and must, by symmetry, also be equivalent to (3), (4),
(5)
and (6).) Now
X = XX*A*
=
XX* A* AY = XAY
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This note was uploaded on 12/29/2011 for the course ECON 505 taught by Professor Penrose during the Spring '11 term at Boğaziçi University.
 Spring '11
 Penrose
 Econometrics

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