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?
Best Answer
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.