Powers of a Matrix

# Powers of a Matrix

Definition: Given a square matrix $A$, for $n$ being a nonnegative integer, $A^n$ is defined as the product matrix taking $A$ and multiplying it by itself $n$-times. If $A$ is invertible, then $A^{-n} = (A^{-1})^{n}$, or the product matrix taking $A^{-1}$ and multiplying it by itself $n$-times. |

For example, consider the following matrix:

(1)\begin{align} A = \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix} \end{align}

To compute $A^2$, take $A$ and multiply it by itself:

(2)\begin{align} A^2 = AA = \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 7 & 10\\ 15 & 22 \end{bmatrix} \end{align}

Additionally, if we wanted to computer $A^{-2}$, we would first need to find $A^{-1} = \begin{bmatrix} -2 & 1\\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$, and then:

(3)\begin{align} A^{-2} = A^{-1}A^{-1} = \begin{bmatrix} -2 & 1\\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} -2 & 1\\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{11}{2} & -\frac{5}{2}\\ -\frac{15}{4} & \frac{7}{4} \end{bmatrix} \end{align}

Theorem 1: If $A$ is a square matrix and let $r$ and $s$ be integers and let $k \in \mathbb{R}$ be a scalar. Then:a) $A^r A^s = A^{r+s}$.b) $(A^r)^s = A^{rs}$.c) If $A^n$ is invertible, then $A^{-n} = (A^{-1})^n$ for $n > 0$.d) If $k ≠ 0$ and the matrix $A$ is invertible, then $(kA)^{-1} = \frac{1}{k}A^{-1}$. |

**Proof of (a):**The matrix $A$ multiplied by itself $r$ times and then multiplied by itself $s$ times is the same as $A$ multiplied by itself $r + s$ times. $\blacksquare$

**Proof of (b):**First take matrix $A$ and multiply it by itself $r$ times. Then take this product and raise it to the $s^{\mathrm{th}}$ power. The total number of times $A$ is multiplied by itself will be $r \cdot s$. $\blacksquare$

**Proof of (c):**Assume that $(A^n)^{-1} = (A^{-1})^n$ is true and multiply both sides on the left by $(A^n)$:

\begin{align} (A^n)(A^n)^{-1} = (A^n)(A^{-1})^n \\ I = \underbrace{AA...AA^{-1}A^{-1}...A^{-1}}_{\mathrm{n-times}} \\ I = I \\ \blacksquare \end{align}

**Proof of (d):**Suppose that $(kA)^{-1} = \frac{1}{k}A^{-1}$. It follows that $(kA)(\frac{1}{k}A^{-1} ) = k \cdot \frac{1}{k} AA^{-1} = I$. Therefore the inverse of $kA$ must be equal to $\frac{1}{k} A^{-1}$. $\blacksquare$