Matrix to the power t.

Question

Compute the matrix $A^t$ for the following cases:

$A_1=\begin{bmatrix}0&0\\0&1\end{bmatrix}, \quad A_2=\begin{bmatrix}-1&0\\0&-2 \end{bmatrix}, \quad A_3=\begin{bmatrix}0&1\\0&0\end{bmatrix}, \quad A_4=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}, \quad A_5=\begin{bmatrix}1&1\\1&1\end{bmatrix}$

And the answers should be:

$ \quad A^{t}_1=\begin{bmatrix}0^t&0\\0&1^t \end{bmatrix}, \quad A^{t}_2=\begin{bmatrix}(-1)^t&0\\0&(-2)^t \end{bmatrix}, \quad A^{t}_3=\begin{bmatrix}\delta(t)&\delta(t-1)\\0&\delta(t) \end{bmatrix},\quad A^{t}_4=\begin{bmatrix}\delta(t)&\delta(t-1)&\delta(t-2)\\0&\delta(t)&\delta(t-1)\\0&0&\delta(t)\end{bmatrix},\quad A^{t}_5=\frac{1}{2}\begin{bmatrix}0^t+2^t&-0^t+2^t\\-0^t+2^t&0^t+2^t \end{bmatrix}$

where $ \delta(j)=0^{|j|}=\left\{ \begin{array}{ll} 1 \ \text{for} \ j = 0\\ 0 \ \text{for} \ j \neq 0\\ \end{array} \right. $

How is this done? What kind of sorcery is this?

I've checked on the net but all I could find were tricky mathematical definitions, while I'm looking for a simple "how to".

Help me Stackexchange, you are my only hope.

Answer 1

For a diagonal $n\times n$ matrix, it is not very hard to check that $$\bigl(D(a_1,\dots,a_n)\bigr)^t=D(a_1^t,\dots,a_n^t).$$

For $A_3$ and $A_4$, just compute the first values to find that $$A_3^2=0,\quad A_4^2=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix},\quad A_4^3=0.$$

For $A_5$, you can prove by induction that $$A_5^t=\begin{pmatrix}2^{t-1}&2^{t-1}\\ 2^{t-1}&2^{t-1}\end{pmatrix}.$$

Answer 2

Raise each matrix to powers and use math induction to prove the given results.

The notation should not bother you if you use your own notation instead of those $\delta$ stuff.

Some like $A_1$ and $A_2$ are obvious and others take some time but are not too bad.