# Power Of Matrices

The matrices can be multiplied to get product matrix and also they demonstrate all other mathematical properties. The power of matrices is another mathematical property of matrix where matrix is raised to a power using an exponent. This brings another question, does the exponent laws applies to matrices or not ? what type of matrices qualifies to be raised to some power ? What about common mathematical identities that involve matrices and power of matrices.

### Exponents or Power of a Number

Exponent or power is a number which tell us how many times a number $base$ should multiplied by itself. If $a$ represents a base and $n$ is its power, then its written as $a^n$ which means

Similarly, a square matrix $A$ and an integer $n$ is given, then $n^th$ power of $A$ is defined as product matrix obtained by multiplying $A$ by itself $n$ times.

$A^n = A \times A \times A \times ... \times A$$A^n = A \times A \times A \times ... \times A$ (n times)

Note that the matrix $A$ is

• a square matrix
• and $A^n$ is a product matrix of same order.

The exponents have their own algebra which is given as follows.

Basic Laws of Exponents

The basic laws of exponents applied to any real number $a \in R$ and these are

• $a^p \times a^q = a^{p + q}$
• $\frac{a^p } {a^q } = a^{p - q}$
• $(a^p)^q = a^{p * q}$
• $(ab)^q = a^{q} * b^{q}$

We need to find out whether these laws applies to square matrices or not. Let us verify this claim with examples.

Example Proof #1

Suppose $A$$A$ is a square matrix of order 2 x 2.

$A = \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}$$A = \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}$

$A^2 = \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix} \times \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix} = \begin{bmatrix}1+4 & 2+6\\2+6 & 4+9\end{bmatrix} = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix}$$A^2 = \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix} \times \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix} = \begin{bmatrix}1+4 & 2+6\\2+6 & 4+9\end{bmatrix} = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix}$

$A^3 = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix} \times \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix} = \begin{bmatrix}5+16 & 10+24\\8+26 & 16+39\end{bmatrix} = \begin{bmatrix}21 & 34\\34 & 55\end{bmatrix}$$A^3 = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix} \times \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix} = \begin{bmatrix}5+16 & 10+24\\8+26 & 16+39\end{bmatrix} = \begin{bmatrix}21 & 34\\34 & 55\end{bmatrix}$

$A^2 \times A^3 = A^{2 + 3} = A^{5}$$A^2 \times A^3 = A^{2 + 3} = A^{5}$

$A^2 \times A^3 = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix} \times \begin{bmatrix}21 & 34\\34 & 55\end{bmatrix} = \begin{bmatrix}105+272 & 170+440\\168+442 & 272+715\end{bmatrix} = \begin{bmatrix}377 & 610\\610 & 987\end{bmatrix}$$A^2 \times A^3 = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix} \times \begin{bmatrix}21 & 34\\34 & 55\end{bmatrix} = \begin{bmatrix}105+272 & 170+440\\168+442 & 272+715\end{bmatrix} = \begin{bmatrix}377 & 610\\610 & 987\end{bmatrix}$

Also,

$A^5 = \begin{bmatrix}377 & 610\\610 & 987\end{bmatrix}$$A^5 = \begin{bmatrix}377 & 610\\610 & 987\end{bmatrix}$

Therefore, both side of the equation is equal.

Example Proof #2

There is not concept of division in matrix, however, you can divide element of matrix by multiplying it with an inverse value which is same as dividing the element. Inverse of a matrix is covered in the next lesson.

If $A$$A$ is a square matrix of order 2 x 2. Then

$A = \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}$$A = \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}$

Therefore,

$\frac{a^p}{a^q}$$\frac{a^p}{a^q}$ is not possible, but if $A$$A$ is an invertible matrix
then

$A.A^{-1} = I$$A.A^{-1} = I$

where

$A^{-1}$$A^{-1}$ is inverse of the matrix $A$$A$ of same order and $I$$I$ is called the identity matrix of same order.

Example Proof #3

The power of power of a matrix is a product matrix with exponents multiplied. If $A$ is a square matrix raised to power $p$ and $A^p$ is also raised to the power $(A^p)^q$, then the resultant product matrix is $A^{p \times q}$ of same order.

Let $A$$A$ be a square matrix with order 2 x 2.

$A = \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}$$A = \begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}$

Then,

$(A^2)^3 = A^{2 \times 3} = A^6$$(A^2)^3 = A^{2 \times 3} = A^6$

$(A^2)^3 = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix}^3$$(A^2)^3 = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix}^3$

$A^{2 \times 3} = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix} \times \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix} \times \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix}$$A^{2 \times 3} = \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix} \times \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix} \times \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix}$

$A^{2 \times 3} = \begin{bmatrix}89 & 144\\144 & 233\end{bmatrix} \times \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix}$$A^{2 \times 3} = \begin{bmatrix}89 & 144\\144 & 233\end{bmatrix} \times \begin{bmatrix}5 & 8\\8 & 13\end{bmatrix}$

$A^{2 \times 3} = \begin{bmatrix}445 + 1152 & 712 + 1872\\720 + 1864 & 1152+3029\end{bmatrix}$$A^{2 \times 3} = \begin{bmatrix}445 + 1152 & 712 + 1872\\720 + 1864 & 1152+3029\end{bmatrix}$

$A^{2 \times 3} = \begin{bmatrix}1597 & 2584\\2584 & 4184\end{bmatrix}$$A^{2 \times 3} = \begin{bmatrix}1597 & 2584\\2584 & 4184\end{bmatrix}$

Also,

$A^6 = \begin{bmatrix}1597 & 2584\\2584 & 4184\end{bmatrix}$$A^6 = \begin{bmatrix}1597 & 2584\\2584 & 4184\end{bmatrix}$

Example Proof #4

The product of defined matrices $AB$ raised to power $p$ is equal to product of powers of individual matrices $A$ and $B$ raised to the power $p$.

$(AB)^p = A^p \times B^p$$(AB)^p = A^p \times B^p$

We can rewrite the equation as

$(AB)^p = (AB)(AB)(AB)... p-times$$(AB)^p = (AB)(AB)(AB)... p-times$

$(AB)^p = A(BA)(BA)B ... p-times$$(AB)^p = A(BA)(BA)B ... p-times$

But we know that $AB$ is not commutative.

$AB \ne BA$$AB \ne BA$

Therefore,

$(AB)^p \ne A^p \times B^p$$(AB)^p \ne A^p \times B^p$

### Why Only Square Matrix ?

Only square matrix is suitable for exponents or to be raised to some powers because of two reasons.

• Non-Square or Singular matrices are not defined. If $A$ is a non-square matrix of order $m \times n$ then $A^p$ is not possible because $m \ne n$ where $m$ is row and $n$ is column of matrix $A$.
• When we need to take inverse which is $A^-1$, the matrix must be a square. Singular matrices are not invertible.

Example Proof #5

Let $A$ be a non-square matrix with order $m \times n$. Let $p$ be an non-negative integer whose value is 2.

$A = \begin{bmatrix}1 & 4 & 2\\2 & 0 & 1\end{bmatrix}$$A = \begin{bmatrix}1 & 4 & 2\\2 & 0 & 1\end{bmatrix}$

Let us try to obtain $A^p = A^2$$A^p = A^2$, then we observe that
it is not possible.

$A_{3 \times 2} \times A_{3 \times 2}$$A_{3 \times 2} \times A_{3 \times 2}$ is undefined where $m = 3$$m = 3$ and $n = 2$$n = 2$. This is because $m \ne n$$m \ne n$ required for multiplication of matrices.

Therefore, a matrix cannot be raised to power unless it is a square matrix.

Example Proof #6

Another reason to use square matrix with power is to find inverse matrix. If $A$ is a matrix that is invertible and we wish to find the inverse matrix $A^{-1}$ such that

$A.A^{-1} = A^{-1}.A = I_{n \times n}$$A.A^{-1} = A^{-1}.A = I_{n \times n}$

The $I_{n \times n}$ represents an identity matrix whose main diagonals are 1 and rest of the entries are 0. It is a square matrix. Therefore, $A$ and $A^{-1}$ must be square matrices.

In the next section, we will explore whether matrices complies with common algebraic identities or not.

### Common Algebraic Identities And Square Matrices

The standard algebraic identities are true for any value of variables. Instead of numbers, we will use square matrix to prove these identities holds for matrices too.

• $(A + B)^2 = A^2 + B^2 + 2AB$
• $A^2 - B^2 = (A + B)(A - B)$
• $(A + B)^3 = A^3 + B^3 + 3A^2B + 3AB^2$
• $A^2B^2 = A(AB)B$

Order Of Multiplication

In matrix multiplication, the order of multiplication is very important because $AB \ne BA$ which is even true for square matrices.

If $AB = BA$, then any one of the following is true.

• The matrix $A = B$ .
• Either $A$ or $B$ is identity matrix $I_n$.
• Either $A$ or $B$ is zero or null matrix $O_n$.
• The $A$ is inverse of $B$ or the matrix $B$ is inverse of $A$.

Let us now verify the common algebraic identities with matrices as variables.

Example Proof #7

We check the following identity : $(A + B )^2 = A^2 + B^2 + 2AB$.

Let $A$$A$ and $B$$B$ be square matrices of order 2 x 2.

$(A + B)^2 = (A + B)(A + B)$$(A + B)^2 = (A + B)(A + B)$

$= A^2 + AB + BA + B^2$$= A^2 + AB + BA + B^2$

But $AB \ne BA$$AB \ne BA$, therefore,

$(A + B)^2$$(A + B)^2$ not possible.

Example Proof #8

We will verify the claim: $A2 - B^2 = (A + B) (A - B)$

Let $A$$A$ and $B$$B$ be square matrix of order 2 x 2.

$A^2 - B^2 = (A + B)(A - B)$$A^2 - B^2 = (A + B)(A - B)$

We can write the right-hand side as

$= A^2 - AB + BA - B^2$$= A^2 - AB + BA - B^2$

But we know that $AB \ne BA$$AB \ne BA$, therefore, above identity is false.

Example Proof #9

We check the identity: $(A + B)^3 = A^3 + B^3 + 3A^2B + 3AB^2$.

Let $A$$A$ and $B$$B$ be square matrices of order 2 x 2.

We multiply (A + B)^3, we get following results

$(A + B)(A + B)(A + B)$$(A + B)(A + B)(A + B)$

Since, we know that $(A + B)^2$$(A + B)^2$ is not possible. Therefore, $(A + B)^3$$(A + B)^3$ is also false.

Example Proof #10

We now verify the identity : $A^2B^2 = A(AB)B$

Let $A$$A$ and $B$$B$ be square matrices and single both are defined. We get following equation from \$latex A^2B^2.

$A^2B^2 = A.A.B.B$$A^2B^2 = A.A.B.B$

We get

$A^2B^2 = A(A.B)B$$A^2B^2 = A(A.B)B$

Therefore, the identity is true because the order of multiplication is same.