Multiplication of Matrices

DEFINITION 1.2.8 (Matrix Multiplication / Product)   Let $A=[a_{ij}]$ be an $m \times n$ matrix and $B= [b_{ij}]$ be an $n \times r$ matrix. The product $A B$ is a matrix $C = [c_{ij}]$ of order $m \times r,$ with

$$c_{ij} = \sum\limits_{k=1}^n a_{ik} b_{kj} = a_{i1} b_{1j} + a_{i2} b_{2j} + \cdots + a_{in} b_{nj}.$$

That is, if $A_{m \times n} = \left[\begin{array}{cccc} & & \cdots & \\ & & \cdots & \\ a_... ...a_{i2} & \cdots & a_{in} \\ & & \cdots & \\ & & \cdots & \\ \end{array}\right]$ and $B_{n \times r} = \left[\begin{array}{ccc} \cdots & b_{1j} & \cdots \\ \cdots ... ... \\ \vdots & \vdots & \vdots \\ \cdots & b_{mj} & \cdots \\ \end{array}\right]$ then

$$A B = [(AB)_{ij}]_{m \times r} {\mbox{ and }} (AB)_{ij} = a_{i1} b_{1j} + a_{i2} b_{2j} + \cdots + a_{in} b_{nj}.$$

Observe that the product $A B$ is defined if and only if
THE NUMBER OF COLUMNS OF MATHEND000# THE NUMBER OF ROWS OF MATHEND000#

For example, if $A= \begin{bmatrix}1 & 2 & 3 \\ 2 & 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix}1 & 2 & 1\\ 0 & 0 & 3 \\ 1 & 0 & 4 \end{bmatrix}$ then

$$A B = \begin{bmatrix}1 +0+ 3 & 2+0+0 & 1 + 6 + 12 \\ 2 + 0+1 & 4+0+0 & 2 + 12 + 4 \end{bmatrix} = \begin{bmatrix}4 & 2 & 19 \\ 3 & 4 & 18 \end{bmatrix}$$

$$= \begin{bmatrix}1 \cdot (1 \ 2 \ 1) + 2 \cdot (0 \ 0 \ 3 ) + 3 \cd... ... 2 \cdot (1 \ 2 \ 1) + 4 \cdot (0 \ 0 \ 3 ) + 1 \cdot (1 \ 0 \ 4) \end{bmatrix}$$ (1.2.1)

$$= \left[\begin{array}{c} \left(\begin{array}{c} 1 \\ 2 \end{array}\... ... 0 + \left(\begin{array}{c} 3 \\ 1 \end{array}\right) \cdot 1\end{array}\right.$$

$$\hspace{.75in} \left(\begin{array}{c} 1 \\ 2 \end{array}\right) \... ...rray}\right) \cdot 0 + \left(\begin{array}{c} 3 \\ 1 \end{array}\right) \cdot 0$$

$$\hspace{1.5in} \left. \begin{array}{c} \left(\begin{array}{c} 1 \... ... + \left(\begin{array}{c} 3 \\ 1 \end{array}\right) \cdot 4 \end{array}\right].$$ (1.2.2)

Observe the following:

1. In this example, while $A B$ is defined, the product $B A$ is not defined.

   However, for square matrices $A$ and $B$ of the same order, both the product $A B$ and $B A$ are defined.

2. The product $A B$ corresponds to operating on the rows of the matrix $B$ (see 1.2.1), and
3. The product $A B$ also corresponds to operating on the columns of the matrix $A$ (see 1.2.2).

DEFINITION 1.2.9   Two square matrices $A$ and $B$ are said to commute if $A B = B A.$

Remark 1.2.10  

1. Note that if $A$ is a square matrix of order $n$ then $A I_n = I_n A.$ Also, a scalar matrix of order $n$ commutes with any square matrix of order $n$ .
2. In general, the matrix product is not commutative. For example, consider the following two matrices $A = \begin{bmatrix}1 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix}1 & 0 \\ 1 & 0 \end{bmatrix}$ . Then check that the matrix product
   $$A B = \begin{bmatrix}2 & 0 \\ 0 & 0 \end{bmatrix} \neq \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = B A.$$

THEOREM 1.2.11   Suppose that the matrices $A, \;B$ and $C$ are so chosen that the matrix multiplications are defined.

1. Then $(A B) C = A (B C).$ That is, the matrix multiplication is associative.
2. For any $k \in {\mathbb{R}}, \; (k A) B = k ( A B) = A ( k B).$
3. Then $A(B + C) = A B + A C.$ That is, multiplication distributes over addition.
4. If $A$ is an $n \times n$ matrix then $A I_n = I_n A = A.$
5. For any square matrix $A$ of order $n$ and $D={\mbox{diag}}(d_1, d_2, \ldots, d_n),$ we have
   • the first row of $D A$ is $d_1$ times the first row of $A;$
   • for $1 \leq i \leq n,$ the $i^{\mbox{th}}$ row of $D A$ is $d_i$ times the $i^{\mbox{th}}$ row of $A.$
   A similar statement holds for the columns of $A$ when $A$ is multiplied on the right by $D.$

Proof. Part 1. $\;\;$ Let $A= [a_{ij}]_{m \times n}, \; B= [b_{ij}]_{n \times p}$ and $C = [c_{ij}]_{p \times q}.$ Then

$$(BC)_{kj} = \sum_{\ell=1}^p b_{k \ell} c_{\ell j} \;\; {\mbox{ and }} \;\; (AB)_{i\ell} = \sum_{k=1}^n a_{i k} b_{k \ell}.$$

Therefore,

$$\bigl( A(BC) \bigr)_{ij} = \sum_{k=1}^n a_{ik} \bigl(BC\bigr)_{kj} = \sum_{k=1}^n a_{ik} \bigl( \sum_{\ell=1}^p b_{k\ell} c_{\ell j}\bigr)$$

$$= \sum_{k=1}^n \sum_{\ell=1}^p a_{ik} \bigl( b_{k\ell}c_{\ell j}\bigr) = \sum_{k=1}^n \sum_{\ell=1}^p \bigl( a_{ik} b_{k\ell} \bigr) c_{\ell j}$$

$$= \sum_{\ell=1}^p \bigl(\sum_{k=1}^n a_{ik} b_{k\ell} \bigr) c_{\ell j} = \sum_{\ell=1}^t \bigl(AB \bigr)_{i\ell}c_{\ell j}$$

$$= \bigl( (AB)C \bigr)_{ij}.$$

Part 5. $\;\;$ For all $j = 1, 2, \ldots, n,$ we have

$$(D A)_{ij} = \sum_{k=1}^n d_{ik} a_{kj} = d_i a_{ij}$$

as $d_{ik} = 0$ whenever $i \neq k.$ Hence, the required result follows.

The reader is required to prove the other parts. height6pt width 6pt depth 0pt

EXERCISE 1.2.12  

1. Let $A$ and $B$ be two matrices. If the matrix addition $A + B$ is defined, then prove that $(A + B)^t = A^t + B^t$ . Also, if the matrix product $A B$ is defined then prove that $(AB)^t = B^t A^t$ .
2. Let $A = [a_1, a_2, \ldots, a_n]$ and $B = \begin{bmatrix}b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}.$ Compute the matrix products $A B$ and $B A.$
3. Let $n$ be a positive integer. Compute $A^n$ for the following matrices:
   $$\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}, \hspace{.25in} \begi... ...hspace{.25in} \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}.$$
   Can you guess a formula for $A^n$ and prove it by induction?
4. Find examples for the following statements.
   1. Suppose that the matrix product $A B$ is defined. Then the product $B A$ need not be defined.
   2. Suppose that the matrix products $A B$ and $B A$ are defined. Then the matrices $A B$ and $B A$ can have different orders.
   3. Suppose that the matrices $A$ and $B$ are square matrices of order $n.$ Then $A B$ and $B A$ may or may not be equal.