$\newcommand{\AA}{\mathbf{A}} \newcommand{\Tr}[1]{\operatorname{Tr}\left[#1\right]}$ I have a problem that I suspect there is a “relatively” simple answer to but it is currently eluding me. I am working with the expansion of a characteristic for a given matrix $\textbf{A}$. Using the notation of exterior algebra we can show that this expansion is given by: $$ \left\vert \lambda\mathbf I_{n\times n} - \mathbf A \right\vert = \sum_{k=0}^n \lambda^{n-k}(-1)^k \Tr{\Lambda^k \mathbf A} =0 $$ The last term in that sum can be calculated by: $$ \Tr{\Lambda^k \mathbf A} = \frac{1}{k!} \begin{vmatrix} \Tr{\AA} & k-1 & 0 & 0 & \ldots & 0 \\ \Tr{\AA^2} & \Tr{\AA} & k-2 & 0 & \ldots & 0 \\ \Tr{\AA^3} & \Tr{\AA^2} & \Tr{\AA} & k-3 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ \Tr{\AA^{k-1}} & \Tr{\AA^{k-2}} & \Tr{\AA^{k-3}} & \ldots & \Tr{\AA} & 1 \\ \Tr{\AA^k} & \Tr{\AA^{k-1}} & \Tr{\AA^{k-2}} & \Tr{\AA^{k-3}} & \ldots & \Tr{\AA} \end{vmatrix} = \frac{\det |\mathbf Q_k|}{k!}$$ where $\mathbf Q_k$ is the matrix inside of the determinant. For some work I am doing I would like to put a condition on the invertability of $\mathbf Q_k$ as a function of $\Tr{\AA}$ and $k$ but unfortunately I cannot figure anything out. Just to note, matrix $\AA$ is $n\times n$ and $1\leq k \leq n$.
Does anyone have any ideas what I could use to determine this? I have tried to calculate the determinant and the smallest eigenvalue/singular value but things just get really messy very quickly.