linear algebra - Determinant of $N times N$ matrix

So the question asks:

For $n \geq 2$, compute the determinant of the following matrix:
$$
B =
\begin{bmatrix}
-X & 1 & 0 & \cdots & 0 & 0 \\
0 & -X & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 & \vdots \\
\vdots & & \ddots & \ddots & 1 & 0 \\
0 & \cdots & \cdots & 0 & -X & 1 \\

a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X)
\end{bmatrix}
$$

Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix:

$\det \begin{bmatrix} -X & 0 \\ 0 & (a_1-X) \end{bmatrix} = -X(a_1-X) - 0 = X^2 - a_1X $

by expansion along the first row:

$\det \begin{bmatrix} -X & 1 & 0 \\ 0 & -X & 0 \\ 0 & 0 & (a_2-X) \end{bmatrix} = (-X) \times\det \begin{bmatrix} -X & 0 \\ 0 & a_2-X \end{bmatrix} - 1 \det\begin{bmatrix} 0 & 0 \\ 0 & a_2-X \end{bmatrix}$

$= (-X)[(-X)(a_2-X) -0] - 0 = X^3 - a_2X^2 $

So it looks like:

$\det \begin{bmatrix}
-X & 1 & 0 & \cdots & 0 & 0 \\
0 & -X & 1 & \ddots & \vdots & \vdots \\

\vdots & \ddots & \ddots & \ddots & 0 & \vdots \\
\vdots & & \ddots & \ddots & 1 & 0 \\
0 & \cdots & \cdots & 0 & -X & 1 \\
a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X)
\end{bmatrix}
= X^{n} - a_{n-1}X^{n-1} - a_{n-2}X^{n-2} ... - a_1X$

Does this look right? Is "prove by induction" valid to use here?

Answer

Let $v= \begin{pmatrix} 1 \\ x \\ x^{2} \\ \cdot \\ \cdot \\ \cdot \\ x^{n-2} \\x^{n-1} \end{pmatrix}$.

Then $Bv = (x-X)v \iff a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} -x^{n} = p(x) =0$. Thus, all the roots $\alpha$ of the monic polynomial $p(x)$ of degree $n$ noted here give us our eigenvectors $v_{\alpha}$, which are linearly independent since they are columns of a Vandermonde matrix. The associated eigenvalues are $\lambda = \alpha - X$.