5.6. BLOCK MULTIPLICATION OF MATRICES 83
Theorem 5.6.2 Let B be a q× p block matrix as in 5.6.19 and let A be a p×n block matrixas in 5.6.20 such that Bis is conformable with As j and each product, BisAs j for s = 1, · · · , pis of the same size so they can be added. Then BA can be obtained as a block matrix suchthat the i jth block is of the form
∑s
BisAs j. (5.6.21)
Proof: From 5.6.18
BisAs j =(
0 Iri×ri 0)
B
0Ips×ps
0
( 0 Ips×ps 0)
A
0Iq j×q j
0
where here it is assumed Bis is ri× ps and As j is ps×q j. The product involves the sth blockin the ith row of blocks for B and the sth block in the jth column of A. Thus there are thesame number of rows above the Ips×ps as there are columns to the left of Ips×ps in those twoinside matrices. Then from Lemma 5.6.1 0
Ips×ps
0
( 0 Ips×ps 0)=
0 0 00 Ips×ps 00 0 0
Since the blocks of small identity matrices do not overlap,
∑s
0 0 00 Ips×ps 00 0 0
=
Ip1×p1 0. . .
0 Ipp×pp
= I
and so∑s
BisAs j =
∑s
(0 Iri×ri 0
)B
0Ips×ps
0
( 0 Ips×ps 0)
A
0Iq j×q j
0
=(
0 Iri×ri 0)
BIA
0Iq j×q j
0
=(
0 Iri×ri 0)
BA
0Iq j×q j
0
Hence the i jth block of BA equals the formal multiplication according to matrix multipli-cation,
∑s
BisAs j.
This proves the theorem.
Example 5.6.3 Let an n×n matrix have the form
A =
(a bc P
)