A relation between some matrix determinants is needed

Question

Let $$C = \begin{bmatrix} B| (\lambda I- A)^{-1}\cdot B| ... |(\lambda I- A)^{-(n-1)} \cdot B\end{bmatrix} $$ an invertible matrix, and let $$ K = \begin{bmatrix}B| A\cdot B| ...|A^{n-1}\cdot B\end{bmatrix}$$ be another invertible matrix. Can anyone find a relation between $det(C)$ and $det(K)$? An equality, inequality, etc. $\lambda \in \mathcal{R}$

Answer 1

If I understand correctly the notation I suppose $B$ is a column vector in the above. We have: $$ (\lambda-A)^{n-1} C = \left[ (\lambda-A)^{n-1} B \mid (\lambda-A)^{n-2}B \mid\ ...\ \mid B \right] $$ Now $(\lambda-A)^{n-1}- (-A)^{n-1}$ is a polynomial of degree $n-2$ in $A$ so the first column may be expanded as a linear combination of the remaining colomns. We may therefore replace the first column by $(-A)^{n-1} B$ when taking determinants. By induction this is true for all columns so we are left with:

$$ \det( (\lambda-A)^{n-1} C) = \det \left[ (\lambda-A)^{n-1} B \mid (\lambda-A)^{n-2}B \mid\ ...\ \mid B \right] = \det \left[(-A)^{n-1} B \mid (-A)^{n-2}B \mid\ ...\ \mid B \right] $$ We then get (I believe the signs cancels completely but you've better check): $$ p_\lambda(A)^{n-1} \det C = \det(K)$$ with $p_A(\lambda)=\det(\lambda-A)$ being the characteristic polynomial of $A$.