Vtyjkyuk

We're going to consider the matrix $beginpmatrix -4 & -15 \ 2 & 7 endpmatrix.$

(a) Let $mathbfP = beginpmatrix 2 & 5 \ 1 & 3 endpmatrix$. Find the $2 times 2$ matrix $mathbfD$ such that
$mathbfP^-1 mathbfD mathbfP = beginpmatrix -4 & -15 \ 2 & 7 endpmatrix.$

(b) Find a formula for $mathbfD^n,$ where $mathbfD$ is the matrix you found in part (a).
(You don't need to prove your answer, but explain how you found it.)

(c) Using parts (a) and (b), find a formula for $beginpmatrix -4 & -15 \ 2 & 7 endpmatrix^n.$

I have completed part a (thanks for the helpful hints), but I am confused on part b. I solved for a few powers of $mathbfD$ to find a pattern. So far I have if $mathbfD = beginpmatrix a & b \ c & d endpmatrix$ then $mathbfD^n = beginpmatrix -4^n & ? \ 0 & -19^n endpmatrix.$ I don't know if I'm solving for this correctly and as you can see, I'm not sure how to find the $b$ part of $mathbfD^n.$ Thanks again!

Let consider

$$beginpmatrix 3 & -5 \ -1 & 2 endpmatrixbeginpmatrix x & y \ z & w endpmatrixbeginpmatrix 2 & 5 \ 1 & 3 endpmatrix=beginpmatrix -4 & -15 \ 2 & 7 endpmatrix\$$
$$iff
D=beginpmatrix x & y \ z & w endpmatrix=beginpmatrix 2 & 5 \ 1 & 3 endpmatrixbeginpmatrix -4 & -15 \ 2 & 7 endpmatrixbeginpmatrix 3 & -5 \ -1 & 2 endpmatrix=$$

$$=beginpmatrix 2 & 5 \ 1 & 3 endpmatrixbeginpmatrix 3 & -10 \-1 & -3 endpmatrix=beginpmatrix -4 & -5 \ 0 & -19 endpmatrix$$

$$A^5 = PDP^-1PDP^-1PDP^-1PDP^-1PDP^-1$$

Can you simplify this expression?

This is a computer aided proof, sage:

It is good to know that sage provides directly the answer symbolically.

sage: A = matrix(QQ, 2, [-4, -15, 2, 7])
sage: A
[ -4 -15]
[ 2 7]
sage: var("n");
sage: A^n
[ -5*2^n + 6 -15*2^n + 15]
[ 2*2^n - 2 6*2^n - 5]

To get this answer, the best way to proceed is to diagonalize the matrix. (We can also triangularize, but let us compare the shape from the $D$ in the OP, some $T$ instead would have been a better notation, and the truly diagonal form below...) Sage diagonalizes for us.

sage: A.jordan_form(transformation=True)
(
[2|0] 
[-+-] [ 1 1]
[0|1], [-2/5 -1/3]
)
sage: Lam, T = A.jordan_form(transformation=True)
sage: T * Lam * T.inverse()
[ -4 -15]
[ 2 7]

The diagonal matrix $Lambda=beginbmatrix2 &\& 1endbmatrix$, Lam above, has an obvious power, which has to be conjugated with the $T$ above to get the answer...