Suppose $t\in \mathbb{C}$, $\Lambda=\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$, what does $t^\Lambda$ mean? Is it $\begin{bmatrix} t^2 & 0 \\ 0 & t \end{bmatrix}$?
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$e^\Lambda=I+\Lambda+\frac12\Lambda^2+\cdots$ – J. W. Tanner Jun 08 '20 at 20:26
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I don’t know what could it be! Maybe we can say t^A= exp(log(t^A)) (whatever this means!!) so we have exp(A)t... – Fabrizio Jun 08 '20 at 20:27
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The easiest way to understand powers of matrices is through the exponential function $f(x)=e^x$. We know that for all $x$: $$ e^x=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots $$ Since taking natural powers of matrices is something we are already (probably) acquainted with, it is almost a "must" to define: $$ e^{\Lambda}=1+\Lambda+\frac{\Lambda^2}{2}+\dots=\sum\limits_{n=0}^{\infty}\frac{\Lambda^n}{n!} $$ Now, if we want to use some base other than $e$ we can use the fact that for all $t\in\mathbb{R}$, we get $t^x=e^{x\ln{(t)}}$ so for matrices: $$ t^{\Lambda}=e^{\ln{(t)}\Lambda}=1+\ln{(t)}\Lambda+\frac{\left(\ln{(t)}\Lambda\right)^2}{2}+\dots $$ Which is usually much more complicated to use. So in your case, the result is correct since if we use the fact that $\Lambda$ is diagonal to compute: $$ e^{\Lambda}=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\Lambda^n=\sum\limits_{n=0}^{\infty}\left(\begin{array}{cc}\frac{2^n}{n!} & 0 \\ 0 & \frac{1}{n!}\end{array}\right)=\left(\begin{array}{cc}\sum\limits_{n=0}^{\infty}\frac{2^n}{n!} & 0 \\ 0 & \sum\limits_{n=0}^{\infty}\frac{1}{n!}\end{array}\right)=\left(\begin{array}{cc}e^2 & 0 \\ 0 & e^1\end{array}\right) $$ And finally since $t^{\Lambda}=e^{\ln{(t)}\Lambda}$ which is also diagonal we get using the same calculation: $$ t^{\Lambda}=\left(\begin{array}{cc}e^{2\ln{(t)}} & 0 \\ 0 & e^{\ln{(t)}}\end{array}\right)=\left(\begin{array}{cc}t^2 & 0 \\ 0 & t\end{array}\right) $$
Ran Kiri
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