Vtyjkyuk

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

$A_1=beginbmatrix0&0\0&1endbmatrix, quad A_2=beginbmatrix-1&0\0&-2 endbmatrix, quad A_3=beginbmatrix0&1\0&0endbmatrix, quad A_4=beginbmatrix0&1&0\0&0&1\0&0&0endbmatrix, quad A_5=beginbmatrix1&1\1&1endbmatrix$

And the answers should be:

$ quad A^t_1=beginbmatrix0^t&0\0&1^t endbmatrix, quad A^t_2=beginbmatrix(-1)^t&0\0&(-2)^t endbmatrix, quad A^t_3=beginbmatrixdelta(t)&delta(t-1)\0&delta(t) endbmatrix,quad A^t_4=beginbmatrixdelta(t)&delta(t-1)&delta(t-2)\0&delta(t)&delta(t-1)\0&0&delta(t)endbmatrix,quad A^t_5=frac12beginbmatrix0^t+2^t&-0^t+2^t\-0^t+2^t&0^t+2^t endbmatrix$

where
$ delta(j)=0^j=left{
beginarrayll
1 textfor j = 0\
0 textfor j neq 0\
endarray
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.

For a diagonal $ntimes n$ 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
$$A_3^2=0,quad A_4^2=beginpmatrix0&0&1\0&0&0\0&0&0endpmatrix,quad A_4^3=0.$$

For $A_5$, you can prove by induction that $$A_5^t=beginpmatrix2^t-1&2^t-1\ 2^t-1&2^t-1endpmatrix.$$

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.