Show that $\det A_n = \sum_{i=0}^{n}a^{2i}$

Question

Let $\mathbb{F}$ be a field and $a\in\mathbb{F}$. For $n \geq 1$ let $$A_n=\begin{pmatrix} 1+a^2 & a & 0 & 0 & \ldots & 0 \\ a & 1+a^2 & a & 0 & \ldots & 0 \\ 0 & a & 1+a^2 & a & \ldots & 0 \\ \vdots & & \ddots & \ddots & \ddots & \vdots \\ 0 & \ldots & \ldots & a & 1+a^2 & a \\ 0 & \ldots & \ldots & 0 & a & 1+a^2 \end{pmatrix}\in M_n(\mathbb{F}).$$

Show that $$\det A_n = \sum_{i=0}^{n}a^{2i}\;\;(1).$$

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My work so far: I'tried a proof by induction and while the basis step is fairly simple, in the induction step I get that (calculating the determinant via the first column) $$\det A_{n+1}=(1+a^2)\cdot \det A_n - a\eta$$ where $\eta$ denotes the determinant that we get by crossing out the first column and second row of $A_n+1$. My problem is that, even if I make use of the induction hypothesis at this step, I don't see how I can arrive at $(1)$.

Answer 1

You already have that

$$\det A_{n+1}=(1+a^2)\cdot \det A_n - a\eta$$ where $\eta$ denotes the determinant that we get by crossing out the first column and second row of $A_{n+1}$.

That is a good start. Developing that determinant $\eta$ along the first row gives $\eta = a \det A_{n-1}$, so that $$ \det A_{n+1}=(1+a^2)\cdot \det A_n - a^2 \det A_{n-1} $$ and you can prove the formula with induction.

The base cases $n=1,2$ can be verified directly (or even simpler, the base cases $n=0, 1$ if you know that the determinant of an empty matrix is $1$).

See also matrix determinant problem on AoPS.