# Matrix to the power of a matrix

If A=$\begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \\ \end{bmatrix}$ and B=$\begin{bmatrix} b & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & b \\ \end{bmatrix}$ then what is $A^B$

I know what A+B is and I know what A*B is. I even know what $e^A$ is but I was wondering what $A^B$ would be. Does it even have any meaning?

$a^b = (e^{\ln a})^b = e^{b \cdot\ln a}$

So I assume that

$A^B = (e^{\ln A})^B = e^{B \cdot\ln A}$

If so then the question becomes what is the natural log of A?

• Do you know what the notation $A^{B}$ means? Have you seen any worked examples? Mar 31, 2018 at 3:40

If $a>0$, then $$\ln{A}:= \begin{bmatrix} \ln{a} & & \\ & \ln{a} & \\ & & \ln{a} \end{bmatrix}.$$ Given that $a^b := e^{b\ln{a}}$, a natural definition is $$A^B:=e^{B\ln{A}},$$ where $$e^X:= \sum_{k=0}^\infty \frac{X^k}{k!}.$$

In this case, since the matrices are diagonal, it follows that $$A^B = e^{B \ln{A}} = \begin{bmatrix} e^{b\ln{a}} & & \\ & e^{b\ln{a}} & \\ & & e^{b\ln{a}} \end{bmatrix} =\begin{bmatrix} a^{b} & & \\ & a^{b} & \\ & & a^{b} \end{bmatrix}.$$