Meaning of power matrix function Suppose $t\in \mathbb{C}$, $\Lambda=\begin{bmatrix}
2 & 0 \\
0 & 1
\end{bmatrix}$, what does $t^\Lambda$ mean? Is it 
$\begin{bmatrix}
t^2 & 0 \\
0 & t
\end{bmatrix}$?
 A: 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:
$$
t^{\Lambda}=\left(\begin{array}{cc}e^{2\ln{(t)}} & 0 \\ 0 & e^{\ln{(t)}}\end{array}\right)=\left(\begin{array}{cc}t^2 & 0 \\ 0 & t\end{array}\right)
$$
