Information Technology Reference
In-Depth Information
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
Search WWH ::




Custom Search