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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?
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 $$ A^B = e^{B \ln{A}} = \begin{bmatrix} e^{b\ln{a}} & & \\ & e^{b\ln{a}} & \\ & & e^{b\ln{a}} \end{bmatrix} =\begin{bmatrix} a^{b} & & \\ & a^{b} & \\ & & a^{b} \end{bmatrix}. $$
