Block Matrices

Let $A$ be an $n \times m$ matrix and $B$ be an $m \times p$ matrix. Suppose $r < m.$ Then, we can decompose the matrices $A$ and $B$ as $A = [P \; Q]$ and $B = \begin{bmatrix}H \\ K \end{bmatrix};$ where $P$ has order $n \times r$ and $H$ has order $r \times p.$ That is, the matrices $P$ and $Q$ are submatrices of $A$ and $P$ consists of the first $r$ columns of $A$ and $Q$ consists of the last $m-r$ columns of $A$ . Similarly, $H$ and $K$ are submatrices of $B$ and $H$ consists of the first $r$ rows of $B$ and $K$ consists of the last $m-r$ rows of $B$ . We now prove the following important theorem.

THEOREM 1.3.5   Let $A = [a_{ij}]= [P \; Q]$ and $B = [b_{ij}]=\begin{bmatrix}H \\ K \end{bmatrix}$ be defined as above. Then

$$A B = P H + Q K.$$

Proof. First note that the matrices $PH$ and $QK$ are each of order $n \times p$ . The matrix products $PH$ and $QK$ are valid as the order of the matrices $P, H, Q$ and $K$ are respectively, $n \times r, \; r \times p, \; n \times (m-r)$ and $(m-r) \times p$ . Let $P= [P_{ij}], \; Q = [Q_{ij}], \; H=[H_{ij}],$ and $K = [k_{ij}]$ . Then, for $1 \le i \le n$ and $1 \le j \le p$ , we have

$$(A B)_{ij} = \sum\limits_{k=1}^m a_{ik} b_{kj} = \sum\limits_{k=1}^r a_{ik} b_{kj} + \sum\limits_{k=r+1}^m a_{ik} b_{kj}$$

$$= \sum\limits_{k=1}^r P_{ik} H_{kj} + \sum\limits_{k=r+1}^m Q_{ik} K_{kj}$$

$$= (PH)_{ij} + (QK)_{ij} = (PH + QK)_{ij}.$$

height6pt width 6pt depth 0pt

Theorem 1.3.5 is very useful due to the following reasons:

1. The order of the matrices $P, Q, H$ and $K$ are smaller than that of $A$ or $B.$
2. It may be possible to block the matrix in such a way that a few blocks are either identity matrices or zero matrices. In this case, it may be easy to handle the matrix product using the block form.
3. Or when we want to prove results using induction, then we may assume the result for $r \times r$ submatrices and then look for $(r+1) \times (r+1)$ submatrices, etc.

For example, if $A = \begin{bmatrix}1 & 2 & 0 \\ 2 & 5 & 0 \end{bmatrix}$ and $B = \begin{bmatrix}a & b \\ c & d \\ e & f \end{bmatrix},$ Then

$$A B = \begin{bmatrix}1& 2 \\ 2& 5 \end{bmatrix} \begin{bmatrix}a&... ...ght] = \begin{bmatrix}a + 2 c & b + 2 d \\ 2 a + 5 c & 2 b + 5 d \end{bmatrix}.$$

If $A= \begin{bmatrix}0 & -1 & 2 \\ 3 & 1 & 4 \\ -2 & 5 & -3 \end{bmatrix},$ then $A$ can be decomposed as follows:

$A = \left[\begin{array}{c\vert cc} 0 & -1 & 2 \\ 3 & 1 & 4 \\ \hline -2 & 5 & -3 \end{array}\right],$ or $\;\;\; A= \left[\begin{array}{cc\vert c} 0 & -1 & 2 \\ 3 & 1 & 4 \\ \hline -2 & 5 & -3 \end{array}\right],$ or

$A=\left[\begin{array}{cc\vert c} 0 & -1 & 2 \\ \hline 3 & 1 & 4 \\ -2 & 5 & -3 \end{array}\right]$ and so on.

Suppose $$A = \begin{array}{rl} & m_1 \;\;\; m_2 \\ \begin{array}{r} n... ...}& \!\! \begin{bmatrix}P & Q \\ R & S \end{bmatrix} \end{array}$$ and $$B = \begin{array}{rl} & s_1 \;\;\; s_2 \\ \begin{array}{c} r... ...}&\!\! \begin{bmatrix}E & F \\ G & H \end{bmatrix} \end{array}.$$ Then the matrices $P, \; Q,\; R, \; S$ and $E, \; F, \; G, \; H,$ are called the blocks of the matrices $A$ and $B,$ respectively.

Even if $A + B$ is defined, the orders of $P$ and $E$ may not be same and hence, we may not be able to add $A$ and $B$ in the block form. But, if $A + B$ and $P + E$ is defined then $A + B = \begin{bmatrix}P + E & Q + F \\ R + G & S + H \end{bmatrix}.$

Similarly, if the product $A B$ is defined, the product $PE$ need not be defined. Therefore, we can talk of matrix product $A B$ as block product of matrices, if both the products $A B$ and $PE$ are defined. And in this case, we have $A B = \begin{bmatrix}P E + Q G & P F + Q H \\ R E + S G & R F + S H \end{bmatrix}.$

That is, once a partition of $A$ is fixed, the partition of $B$ has to be properly chosen for purposes of block addition or multiplication.