Advice on proof in linear algebra. I just wrote my first proof in linear algebra so I'd love some advice on the things that go well and what could be improved upon. It's a proof by induction.
Theorem:
Let $A_n$ be a $n\times n$ matrix of the form:
$\begin{pmatrix}
2 & 1 & 0 & 0 && & \cdots & 0\\
1 & 2 & 1 & 0 && & \cdots & 0\\
0 & 1 & 2 & 1 && & \cdots & 0\\
0 & 0 & 1 & 2 && & \cdots & 0\\
\vdots& & & & \ddots & && \vdots\\
0&\cdots&&&&2&1&0\\
0&\cdots&&&&1&2&1\\
0&\cdots&&&&0&1&2\\
\end{pmatrix}$
Then $det(A_n)=3(n-1)$.
Proof:
We'll give a proof by mathematical induction.
Let $n=2$:
$\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}=4-1=3$.
Let $n=3$:
$\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}-\begin{vmatrix}
1&0\\
1&2\\
\end{vmatrix}=8-2=6$
Let $n=4$:
$\begin{vmatrix}
2&1&0&0\\
1&2&1&0\\
0&1&2&1\\
0&0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}-\begin{vmatrix}
1&0&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}-\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}=12-3=9$
Notice how due to the repetitive nature of our matrix, we'll candevise a recursive formula for our determinant: $|A_n|=2|A_{n-1}|-|A_{n-2}|$.
If we assume: $|A_n|=3(n-1)$, $|A_{n-1}|=3(n-2)$, $|A_{n-2}|=3(n-3)$, then:
$|A_n|=6(n-2)-3(n-3)=3(n-1)$
Since we checked $n=2, 3$; we conclude that $|A_n|=3(n-1)$ for all $n \in \mathbb{N} | n>1$

Let $d_n = \det A_n$. Note that $d_1 = 2, d_2 = 3$.
For the induction step, look at the (clunky) picture below:

Note that $d_n = 2 d_{n-1} -d_{n-2}$.
Now show that $n \mapsto 3(n-1)$ satisfies the equation with the same
initial conditions.
