Let $A = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)$. Prove for $n \geq 1$ using induction that $A^n =$ ... Can someone check to see if my proof is correct? Feel free to nitpick, trying to get better at writing proofs. Here's the problem:
Let $A = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)$. Prove for $n \geq 1$ that $A^n = 4^n \bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^n \bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)$ by induction.
Note: $A^n$ means $A$ matrix multiplied by itself $n$ times. So for example, $A^2 = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)\bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr) = \bigl( \begin{smallmatrix}2 & 14 \\ -7 & 23\end{smallmatrix}\bigr)$
My attempt at the proof:
Base case, $n=1:$ $A^1 = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr) = 4^1\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr)+ 3^1\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr) \ \checkmark$
Assume the inductive hypothesis, $n=k: A^k = 4^k\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^k\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)$
WTS that the proposition holds for $n=k+1: A^{k+1} = 4^{k+1}\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^{k+1}\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)$
\begin{align*}
A^k A = \left(4^k\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^k\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)
\right)\bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)
&= 4^k\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr)\bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr) + 3^k\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)\bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr) \\
&= 4^k\bigl( \begin{smallmatrix}-4 & 8 \\ -4 & 8\end{smallmatrix}\bigr)
+ 3^k\bigl( \begin{smallmatrix}6 & -6 \\ 3 & -3\end{smallmatrix}\bigr) \\
&= (4^k)(4)\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr)  + 
(3^k)(3)\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr) \\
A^k A &= 4^{k+1}\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^{k+1}\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr) \\
A^k A &= A^{k+1}
\end{align*}
By induction the proposition holds for all $n \geq 1$
 A: The proof looks perfect mathematically. Stylistically, it looks pretty good as well, although since you mentioned that you'd like some nitpicking, here goes:
Depending on the context in which you need to write this proof, the phrasing of the induction could be improved a little bit. (If this is for a class, and you've been taught to write induction arguments in a very particular manner, then of course you do not want to mess with it.) In particular, proofs are generally easier to read when written in sentence form, and so the flow of the argument might be improved by adding a few more words. What follows are the words I'd be likely to use, if writing the proof in a very detailed manner. This is of course not the only way to phrase the argument, and as you write more proofs you'll develop your own mathematical "tone." (In particular, you may read this as a bit overkill.)

We show the result by induction on $n$. For the base case of $n=1$, we have
$$\text{[computation]}$$.
Now, we show the inductive step. Assume that the proposition holds for some integer $n=k$, i.e. that [statement]. We wish to show that the proposition holds for $n=k+1$, i.e. that [statement]. Indeed,
$$\text{[computation]},$$
where we have used the inductive hypothesis in the first step. We have thus proven the proposition for all $n\geq 1$ by induction.

I'd also recommend moving the note that $A^kA=A^{k+1}$ to the beginning of the inductive-step computation, so that the computation reads
\begin{align*}
A^{k+1}&=A^kA\\
&=\ldots\\
&=4^{k+1}\bigl(\begin{smallmatrix}-1&2\\-1&2\end{smallmatrix}\bigr)+3^{k+1}\bigl(\begin{smallmatrix}2&-2\\1&-1\end{smallmatrix}\bigr)
\end{align*}
and the reader can see directly, just from glancing at the computation, that it is a sequence of steps verifying exactly what you want it to.
(Lastly, I'd also use \begin{pmatrix}\end{pmatrix} over \bigl\begin{smallmatrix}\end{smallmatrix}\bigr -- it's not like you have to pay for space and I think it looks less cluttered -- but this is purely a personal preference.)
