Algebraic Properties of Matrix Operations

In this page, we give some general results about the three operations: addition, multiplication, and multiplication with numbers, called scalar multiplication.

From now on, we will not write (mxn) but mxn.

Properties involving Addition. Let A, B, and C be mxn matrices. We have

1.

A+B = B+A

2.

(A+B)+C = A + (B+C)

3.

$A + {\cal O} = A$
where $\cal O$ is the mxn zero-matrix (all its entries are equal to 0);

4.

$A+B = {\cal O}$ if and only if B = -A.

Properties involving Multiplication.

1.

Let A, B, and C be three matrices. If you can perform the products AB, (AB)C, BC, and A(BC), then we have

(AB)C = A (BC)

Note, for example, that if A is 2x3, B is 3x3, and C is 3x1, then the above products are possible (in this case, (AB)C is 2x1 matrix).

2.

If $\alpha$ and $\beta$ are numbers, and A is a matrix, then we have

$$\alpha (\beta A) = (\alpha \beta) A$$

3.

If $\alpha$ is a number, and A and B are two matrices such that the product $A\cdot B$ is possible, then we have

$$\alpha (AB) = (\alpha A)B = A (\alpha B)$$

4.

If A is an nxm matrix and $\cal O$ the mxk zero-matrix, then

$$A {\cal O} = {\cal O}$$

Note that $A {\cal O}$ is the nxk zero-matrix. So if n is different from m, the two zero-matrices are different.

Properties involving Addition and Multiplication.

1.

Let A, B, and C be three matrices. If you can perform the appropriate products, then we have

(A+B)C = AC + BC

and

A(B+C) = AB + AC

2.

If $\alpha$ and $\beta$ are numbers, A and B are matrices, then we have

$$\alpha (A+B) = \alpha A + \alpha B$$

and

$$(\alpha +\beta)A = \alpha A + \beta B$$

Example. Consider the matrices

$$A = \left(\begin{array}{cc} 0&1\\ -1&0\\ \end{array}\right)... ...nd}\; C = \left(\begin{array}{ccc} 0&1&5\\ \end{array}\right).$$

Evaluate (AB)C and A(BC). Check that you get the same matrix.

Answer. We have

$$AB = \left(\begin{array}{c} -1\\ -2\\ \end{array}\right)$$

so

$$(AB)C = \left(\begin{array}{c} -1\\ -2\\ \end{array}\right)... ...t(\begin{array}{ccc} 0&-1&-5\\ 0&-2&-10\\ \end{array}\right).$$

On the other hand, we have

$$BC = \left(\begin{array}{ccc} 0&2&10\\ 0&-1&-5\\ \end{array}\right)$$

so

$$A(BC) = \left(\begin{array}{cc} 0&1\\ -1&0\\ \end{array}\ri... ...t(\begin{array}{ccc} 0&-1&-5\\ 0&-2&-10\\ \end{array}\right).$$

Example. Consider the matrices

$$X = \left(\begin{array}{c} a\\ b\\ c\\ \end{array}\right)\... ...ray}{cccc} \alpha & \beta & \nu & \gamma\\ \end{array}\right).$$

It is easy to check that

$$X = a \left(\begin{array}{c} 1\\ 0\\ 0\\ \end{array}\right... ...) + c \left(\begin{array}{c} 0\\ 0\\ 1\\ \end{array}\right)$$

and

$$Y = \alpha \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ \end{ar... ...ma \left(\begin{array}{cccc} 0 &0 & 0& 1\\ \end{array}\right).$$

These two formulas are called linear combinations. More on linear combinations will be discussed on a different page.

We have seen that matrix multiplication is different from normal multiplication (between numbers). Are there some similarities? For example, is there a matrix which plays a similar role as the number 1? The answer is yes. Indeed, consider the nxn matrix

$$I_n = \left(\begin{array}{ccccc} 1&0&0&\cdots&0\\ 0&1&0&\cdo... ...cdot\\ \cdot&&&&\cdot\\ 0&0&0&\cdots&1\\ \end{array}\right).$$

In particular, we have

$$I_2 = \left(\begin{array}{ccc} 1&0\\ 0&1\\ \end{array}\righ... ...egin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{array}\right).$$

The matrix I_n has similar behavior as the number 1. Indeed, for any nxn matrix A, we have

A I_n = I_n A = A

The matrix I_n is called the Identity Matrix of order n.

Example. Consider the matrices

$$A = \left(\begin{array}{cc} 1&2\\ -1&-1\\ \end{array}\right... ...B = \left(\begin{array}{cc} -1&-2\\ 1&1\\ \end{array}\right).$$

Then it is easy to check that

$$AB = I_2 \;\;\mbox{and}\;\; BA = I_2.$$

The identity matrix behaves like the number 1 not only among the matrices of the form nxn. Indeed, for any nxm matrix A, we have

$$I_n A = A\;\;\mbox{and}\;\; A I_m = A.$$

In particular, we have

$$I_4 \left(\begin{array}{c} a\\ b\\ c\\ d\\ \end{array}\ri... ... \left(\begin{array}{c} a\\ b\\ c\\ d\\ \end{array}\right).$$

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