Now what if instead of numbers, the entries, A,B,C,D,E,F,G are matrices of a size such that the
multiplications and additions needed in the above formula all make sense. Would the formula be true in
this case? I will show below that this is true.
Suppose A is a matrix of the form
( A ⋅⋅⋅ A )
| 1.1 . 1.m |
A = |( .. .. .. |) (4.15)
Ar1 ⋅⋅⋅ Arm
(4.15)
where A_{ij} is a s_{i}× p_{j} matrix where s_{i} is constant for j = 1,
⋅⋅⋅
,m for each i = 1,
⋅⋅⋅
,r. Such a matrix is
called ablock matrix, also a partitioned matrix. How do you get the block A_{ij}? Here is how for A an
m × n matrix:
In the block column matrix on the right, you need to have c_{j}− 1 rows of zeros above the small p_{j}× p_{j}
identity matrix where the columns of A involved in A_{ij} are c_{j},
⋅⋅⋅
,c_{j} + p_{j}− 1 and in the block row matrix
on the left, you need to have r_{i}− 1 columns of zeros to the left of the s_{i}×s_{i} identity matrix where the rows
of A involved in A_{ij} are r_{i},
⋅⋅⋅
,r_{i} + s_{i}. An important observation to make is that the matrix on the right
specifies columns to use in the block and the one on the left specifies the rows used. Thus the block
A_{ij} in this case is a matrix of size s_{i}× p_{j}. There is no overlap between the blocks of A. Thus
the identity n × n identity matrix corresponding to multiplication on the right of A is of the
form
( I 0 )
| p1×p1 . |
|( .. |)
0 Ipm×pm
these little identity matrices don’t overlap. A similar conclusion follows from consideration of the matrices
I_{si×si}.
Next consider the question of multiplication of two block matrices. Let B be a block matrix of the
form
and that for all i,j, it makes sense to multiply B_{is}A_{sj} for all s ∈
{1,⋅⋅⋅,p}
. (That is the two matrices, B_{is}
and A_{sj} are conformable.) and that for fixed ij, it follows B_{is}A_{sj} is the same size for each s so that it
makes sense to write ∑_{s}B_{is}A_{sj}.
The following theorem says essentially that when you take the product of two matrices, you can do it
two ways. One way is to simply multiply them forming BA. The other way is to partition both matrices,
formally multiply the blocks to get another block matrix and this one will be BA partitioned.
Before presenting this theorem, here is a simple lemma which is really a special case of the
theorem.
Lemma 4.4.1Consider the following product.
( )
0 ( )
|( I |) 0 I 0
0
where the first is n×r and the second is r ×n. The small identity matrix I is an r ×r matrix and there arel zero rows above I and l zero columns to the left of I in the right matrix. Then the product of thesematrices is a block matrix of the form
( )
| 0 0 0 |
( 0 I 0 )
0 0 0
Proof:From the definition of the way you multiply matrices, the product is
(( ) ( ) ( ) ( ) ( ) ( ) )
|| 0 | | 0 | | 0 | | 0 | | 0 | | 0 | |
(( I ) 0⋅⋅⋅( I ) 0( I ) e1⋅⋅⋅( I ) er( I ) 0⋅⋅⋅( I ) 0 )
0 0 0 0 0 0
which yields the claimed result. In the formula e_{j} refers to the column vector of length r which has a 1 in
the j^{th} position. ■
Theorem 4.4.2Let B be a q × p block matrix as in 4.17and let A be a p × n block matrix as in 4.18such that B_{is}is conformable with A_{sj}and each product, B_{is}A_{sj}for s = 1,
⋅⋅⋅
,p is of the same size so theycan be added. Then BA can be obtained as a block matrix such that the ij^{th}block is of theform
where here it is assumed B_{is} is r_{i}×p_{s} and A_{sj} is p_{s}×q_{j}. The product involves the s^{th} block in the i^{th} row
of blocks for B and the s^{th} block in the j^{th} column of A. Thus there are the same number of rows above
the I_{ps×ps} as there are columns to the left of I_{ps×ps} in those two inside matrices. Then from Lemma
4.4.1
which equals the ij^{th} block of BA. Hence the ij^{th} block of BA equals the formal multiplication according
to matrix multiplication, ∑_{s}B_{is}A_{sj}.■
Example 4.4.3Let an n × n matrix have the form
( )
A = a b
c P
where P is n − 1 × n − 1. Multiply it by
( )
p q
B = r Q
where B is also an n × n matrix and Q is n − 1 × n − 1.
You use block multiplication
( )( ) ( )
a b p q = ap+ br aq + bQ
c P r Q pc+ Pr cq + PQ
Note that this all makes sense. For example, b = 1 × n − 1 and r = n − 1 × 1 so br is a 1 × 1. Similar
considerations apply to the other blocks.
Here is a very significant application. A matrix is called block diagonal if it has all zeros except for
square blocks down the diagonal. That is, it is of the form
( )
A1 0
|| A2 ||
A = || .. ||
( . )
0 Am
where A_{j} is a r_{j}× r_{j} matrix whose main diagonal lies on the main diagonal of A. Then by block
multiplication, if p ∈ ℕ the positive integers,
( )
Ap1 0
|| Ap ||
Ap = || 2 . || (4.20)
( .. )
0 Apm
(4.20)
Also, A^{−1} exists if and only if each block is invertible and in fact, A^{−1} is given by the above when
p = −1.