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We're going to consider the matrix $\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$

(a) Let $\mathbf{P} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Find the $2 \times 2$ matrix $\mathbf{D}$ such that $\mathbf{P}^{-1} \mathbf{D} \mathbf{P} = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$

(b) Find a formula for $\mathbf{D}^n,$ where $\mathbf{D}$ 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 $\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^n.$

I have completed part a (thanks for the helpful hints), but I am confused on part b. I solved for a few powers of $\mathbf{D}$ to find a pattern. So far I have if $\mathbf{D} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ then $\mathbf{D}^{n} = \begin{pmatrix} -4^n & ? \\ 0 & -19^n \end{pmatrix}.$ 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 $\mathbf{D}^{n}.$ Thanks again!

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    $\begingroup$ If we call your original matrix $A,$ the definition $P^{-1}DP = A$ also says $PAP^{-1} = D$ $\endgroup$ – Will Jagy Sep 1 '18 at 1:07
  • $\begingroup$ write $D^3$ in terms of $P$, $P^{-1}$, and $A$. Do you notice any convenient patterns? Hint: try just literally using the letters P, D, and A. don't worry about the specific matrices. $\endgroup$ – The Count Sep 1 '18 at 1:34
  • $\begingroup$ If you’ve done this the way it’s intended, $b$ should be $0$. $\endgroup$ – amd Sep 1 '18 at 4:02
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Let consider

$$\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x & y \\ z & w \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}=\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\\$$ $$\iff D=\begin{pmatrix} x & y \\ z & w \end{pmatrix}=\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}=$$

$$=\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 3 & -10 \\-1 & -3 \end{pmatrix}=\begin{pmatrix} -4 & -5 \\ 0 & -19 \end{pmatrix}$$

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

Can you simplify this expression?

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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 $\Lambda=\begin{bmatrix}2 &\\& 1\end{bmatrix}$, Lam above, has an obvious power, which has to be conjugated with the $T$ above to get the answer...

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