I'm having a tough time proving statements that involve the determinant of an arbitrary matrix and was wondering if there is just an easier way to compute it or some simpler equivalent definition for it. Just to give an example:
Let $K$ be a field, let $\lambda,a_0,a_1 \dots a_n \in K$ and $n \in \mathbb{N}$. Prove that $$
A= \begin {pmatrix} \lambda &0 &0 &\dots & a_0 \\ -1 & \lambda &0 & \dots &a_1 \\0 &-1&\lambda &\dots& a_2 \\ \vdots & \dots & \dots & \dots &a_n\end{pmatrix}\in K^{(n+1)\times(n+1)}, \det(A)= a_n\lambda^n+\dots+ a_2\lambda^2+a_1\lambda+a_0$$
I have zero clue how I can go about this and the definition of the determinant is not very intuitive . My best guess was using the Laplace expansion so that $n+1 \times n+1$ matrix is not worrysome so that I can deal with the submatrix but I'm not sure where I can go from there. I would appreciate tips on how I should approach these proofs. (Also I would appreciate some help regarding the example problem as well).
Best Answer
"Arbitrary matrix" is misleading, and the problem you have isn't as hard.
Just expand the matrix along the first row,then $$\det = \lambda \det (\text{the Right Down} (n-1)\times (n-1) \text{ minor}) + (-1)^{n+2}a_0\cdot (\text{An Upper Triangular}).$$
$\text{The right down} (n-1)\times (n-1) \text{minor}$ can be computed by induction.