1.2. BLOCK MULTIPLICATION OF MATRICES 11

where the first is n× r and the second is r×n. The small identity matrix I is an r× r matrixand there are l zero rows above I and l zero columns to the left of I in the right matrix.Then the product of these matrices is a block matrix of the form 0 0 0

0 I 00 0 0

 .

Proof: From the definition of matrix multiplication, the product is  0I0

0 · · ·

 0I0

e1 · · ·

 0I0

er · · ·

 0I0

0

which yields the claimed result. In the formula e j refers to the column vector of length rwhich has a 1 in the jth position. This proves the lemma. ■

Theorem 1.2.2 Let B be a q× p block matrix as in (1.3) and let A be a p×n blockmatrix as in (1.4) such that Bis is conformable with As j and each product, BisAs j for s =1, · · · , p is of the same size, so that they can be added. Then BA can be obtained as a blockmatrix such that the i jth block is of the form

∑s

BisAs j. (1.5)

Proof: From (1.2)

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 1.2.1 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

