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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?

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    $\begingroup$ Your $2\times 2$ matrix must be $\begin{bmatrix} -X & 1 \\ a_0 & (a_1-X) \end{bmatrix}$. You have also done the same mistake for the next case. I do not think you can use induction here, but you can get an intuitive idea about the general expression. $\endgroup$
    – Rajat
    Apr 8, 2016 at 4:08
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    $\begingroup$ That's essentially the companion matrix. The induction step that works is expansion by minors along the first column. $\endgroup$
    – user5713492
    Apr 8, 2016 at 4:29

2 Answers 2

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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 $\lambda = \alpha - X$.

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  • $\begingroup$ Thanks! Since the characteristic polynomial consists the trace and the determinant. So the detminant of this Vandermonde matrix is $a_0$. Since B is similar to the matrix, so det(B) = $a_0$. Am I right? $\endgroup$
    – Betty
    Apr 8, 2016 at 11:42
  • $\begingroup$ You are not right. You've already conjectured correctly what the determinant of $B$ should be: a polynomial in $X$. For example in your case for $n=2$, you found out that $\det B = X^{2} - a_{1} X - a_{0}$. In my answer I found linearly independent eigenvectors with associated eigenvalues. Determinant is the product of eigenvalues. Let $n = 2$ and $\alpha_{1}, \alpha_{2}$ are the roots of $p(x) = x^{2} - a_{1}x - a_{0}$, then by my answer above $\det B = (\alpha_{1} - X)(\alpha_{2} -X) = X^{2} - (\alpha_{1} +\alpha_{2})X + \alpha_{1} \alpha_{2}$ - roots and coefficients are related (how?) $\endgroup$
    – Daniel Akech Thiong
    Apr 8, 2016 at 13:14
  • $\begingroup$ Note that this proof requires a bit of elementary algebraic geometry in order to work. Per se, the roots of the polynomial $p$ may fail to be distinct, in which case you do not get $n$ linearly independent eigenvectors. So you need to first WLOG assume that the resultant of the polynomial $p$ and its derivative $p'$ is $\neq 0$ (this is true on a Zariski-dense subset of the space of all polynomials, so you can WLOG assume it). $\endgroup$
    – darij grinberg
    Apr 11, 2019 at 18:55
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To add a final touch somewhat as a synthesis of the (thorough) answers of @user5713492 and @akech: the global result is that the companion matrix of polynomial p(X) is diagonalized with a Vandermonde matrix V(r_1,r_2,\cdots r_n) where the $r_k$ are the roots of p(X)" see https://en.wikipedia.org/wiki/Vandermonde_matrix

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