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The first and last matrix in this product are invertible. If
S
is also invertible, the middle
matrix is invertible, as is the matrix
M
itself. The inverse of
M
,
M
−
1
,isgivenbytheproduct
M
−
1
=
I
M
−
1
1,1
M
−
1
−
1,1
M
1,2
I
0
0
(6.9)
S
−
1
M
2,1
M
−
1
I
−
I
0
0
1,1
This follows from three observations: a) the inverse of a product is the product of the inverses
in reverse order (see Lemma
6.2.1
), b) the inverse of a 2
2 upper (lower) triangular matrix
is the matrix with the off-diagonal term negated, and c) the inverse of a 2
×
2diagonalmatrix
is a diagonal matrix in which the
i
th diagonal element is the multiplicative inverse of the
i
th
diagonal element of the original matrix. (See Problem
6.13
for the latter two results.)
The following fact is useful in inverting SPD matrices.
×
LEMMA
6.5.3
If
M
is an
n × n
SPD matrix, its Schur complement is also SPD.
Proof
Represent
M
as shown in (
6.7
). In (
6.6
)let
x
=
u
·
v
;thatis,let
x
be the concate-
nation of the two column vectors. Then
M
1,1
u
+
M
1,2
v
M
2,1
u
+
M
2,2
v
x
T
M
x
=
u
T
,
v
T
=
u
T
M
1,1
u
+
u
T
M
1,2
v
+
v
T
M
2,1
u
+
v
T
M
2,2
v
If we say that
M
−
1
u
=
−
1,1
M
1,2
v
and use the fact that
M
1,2
=
M
2,1
and
M
−
1
1,1
T
=
M
1,1
−
1
=
M
−
1
1,1
,itisstraightforward
to show that
S
is symmetric and
x
T
M
x
=
v
T
S
v
where
S
is the Schur complement of
M
.Thus,if
M
is SPD, so is its Schur complement.
6.5.3 Inversion of Triangular Matrices
Let
T
be
n
n
lower triangular and non-singular. Without loss of generality, assume that
n
=
2
r
.(
T
can be extended to a 2
r
×
2
r
matrix by placing it on the diagonal of a 2
r
2
r
×
×
matrix along with a 2
r
2
r
−
n
×
−
n
identity matrix.) Represent
T
as a 2
×
2matrixof
n/
2
×
n/
2matrices:
T
=
T
1,1
0
T
2,1
T
2,2
The inverse of
T
, which is lower triangular, is given below, as can be verified directly:
T
−
1
=
T
−
1
1,1
0
T
−
1
2,2
T
2,1
T
−
1
T
−
1
2,2
−
1,1
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