Determinant of a companion matrix

Question

I have to find determinant of $$A := \begin{bmatrix}0 & 0 & 0 & ... &0 & a_0 \\ -1 & 0 & 0 & ... &0 & a_1\\ 0 & -1 & 0 & ... &0 & a_2 \\ 0 & 0 & -1 & ... &0 & a_3 \\ \vdots &\vdots &\vdots & \ddots &\vdots&\vdots \\0 & 0 & 0 & ... &-1 & a_{n-1} \end{bmatrix} + t I_{n \times n}$$

It is not a difficult thing to do. My method is as follows :

$$\begin{bmatrix}0 & 0 & 0 & ... &0 & a_0 \\ -1 & 0 & 0 & ... &0 & a_1\\ 0 & -1 & 0 & ... &0 & a_2 \\ 0 & 0 & -1 & ... &0 & a_3 \\ \vdots &\vdots &\vdots & \ddots &\vdots&\vdots \\0 & 0 & 0 & ... &-1 & a_{n-1} \end{bmatrix} + t I_{n \times n} = \begin{bmatrix}t & 0 & 0 & ... &0 & a_0 \\ -1 & t & 0 & ... &0 & a_1\\ 0 & -1 & t & ... &0 & a_2 \\ 0 & 0 & -1 & ... &0 & a_3 \\ \vdots &\vdots &\vdots & \ddots &\vdots&\vdots \\0 & 0 & 0 & ... &-1 & a_{n-1} + t \end{bmatrix} $$

Performing the row reduction of type $R_{k+1} \to R_{k+1} + \dfrac{1}{t}R_k$

I get an upper triangular matrix

$$\begin{bmatrix}t & 0 & 0 & ... &0 & a_0 \\ 0 & t & 0 & ... &0 & a_1 + \dfrac {a_0} t\\ 0 & 0 & t & ... &0 & a_2 + \dfrac{a_1}{t} + \dfrac {a_0} {t^2} \\ 0 & 0 & 0 & ... &0 & a_3 + \dfrac{a_2}{t} + \dfrac{a_1}{t^2} + \dfrac {a_0} {t^3} \\ \vdots &\vdots &\vdots & \ddots &\vdots&\vdots \\0 & 0 & 0 & ... &0 & a_{n-1} + t + \sum_{k=0}^{n-2} \dfrac{a_{k}}{t^{(n-1) - k }} \end{bmatrix} $$

Determinant of which is $t^n + \sum^{n-1}_{k = 1} a_k t^{k}$.

My friend says this is not a rigorous proof and that I have to use induction to prove $$\det A = t^n + \sum^{n-1}_{k = 1} a_k t^{k}$$ She says that I have only found a formula for $\det A$ and I can't be sure if it works for all $n\in \Bbb N$ without a proof. Is she correct?

Answer 1

Yes, the method is completely rigorous, because here $t$ is, algebraically, an indeterminate.

If the coefficients of the matrix are supposed to be in a field $F$, then the computation you make takes place in the field $F(t)$ of rational functions in the indeterminate $t$. No problem here in considering $t^{-1}$, because $t$ is a nonzero element of the field.

To be picky, induction should be needed, but laying out the argument is very easy.

Answer 2

The argument can be made rigorous by the following identity: let $U$ be the final triangular matrix and let $D$ be the matrix containing only a subdiagonal of $1, \cdots, 1$. Then one has

$$ \left(I - \frac{1}{t} D\right) U = (A + t I) $$ hence $\det(A+tI)= \det(I - \frac{1}{t} D) \det (U) = \det(U)$

It means that this process is actually an $LU$ decomposition of $A+tI$.

For a complete calculation, let

$$\renewcommand{\arraystretch}{2} \begin{array}{rcl}{P}_{0}&=&0\\ {P}_{i}&=&\displaystyle \sum _{p = 0}^{i-1} {a}_{p} {t}^{p} \quad i \geqslant 1\\ L&=&{\left({{\delta}}_{i}^{j}-{t}^{{-1}} {{\delta}}_{i}^{j+1}\right)}_{i , j}\\ U&=&{\left(t {{\delta}}_{i}^{j}+{t}^{1-i} {P}_{i} {{\delta}}_{j}^{n}\right)}_{i , j} \end{array}$$

Then, using ${P}_{i}-{P}_{i-1} = {a}_{i-1} {t}^{i-1}$ for $i \geqslant 1$,

$$\renewcommand{\arraystretch}{2} \begin{array}{rcl}{\left(L U\right)}_{i , j}&=&\displaystyle \sum _{k = 1}^{n} \left({{\delta}}_{i}^{k}-{t}^{{-1}} {{\delta}}_{i}^{k+1}\right) \left(t {{\delta}}_{k}^{i}+{t}^{1-k} {P}_{k} {{\delta}}_{j}^{n}\right)\\ &=&t {{\delta}}_{i}^{j}+{t}^{1-i} {P}_{i} {{\delta}}_{j}^{n}-{{\delta}}_{i}^{j+1}-{t}^{1-i} {P}_{i-1} {{\delta}}_{j}^{n}\\ &=&t {{\delta}}_{i}^{j}-{{\delta}}_{i}^{j+1}+{a}_{j-1} {{\delta}}_{j}^{n}\\ &=&{\left(A+t I\right)}_{i , j} \end{array}$$

where $\delta_i^j$ is Kronecker's delta.

Answer 3

$-A$ is the companion matrix for the polynomial $$p(x) = a_0 + a_1 x + \dots + a_{n-1} x^{n-1} + x^n.$$ Hence the eigenvalues of $-A$ are the roots $r_k$ of this polynomial. Since adding a multiple of the identity to a matrix just shifts eigenvalues, the eigenvalues of $A + tI$ are the quantities $t-r_k$. Hence, since the determinant of a matrix is the product of its eigenvalues, we have $$\det(A + tI) = (t - r_1)(t - r_2) \dots (t-r_n) = p(t)$$ as required.