Request for literature on equality $\det{(s^2 I_n + A s + B)} = 0$

Question

We know the roots of equality $\det(s I_n - A) = 0$, such that variable $s \in \mathbb{C}$ are eigenvalues of matrix $A$. Do we know anything about equality $\det{(s^2I_n + A_1 s + A_0)} = 0$.

I try to understand the eigenvalues of matrix $S = \begin{bmatrix} A & B \\ G & H \end{bmatrix}_{(n+m)\times(n+m)}$ with Schur identity, below:

$$\det{(sI_{n+m}-S)} = \det{(sI_m-H)} \det{(sI_n-A - B \, (sI_m-H)^{-1}) G)}$$

On the case $G=F$ and $H = -I_m$, the determinant $\det{(S)}$ follows as below

$ \begin{align*} \det{(sI_m-I_m)} \det{(sI_n - A - B \, (s \, I_m - H)^{-1} \, G)} & = (s+1)^m \det{\left(sI_n-A - B \, \frac{I_m}{s+1} G\right)} \\ & = (s+1)^{m-n} \det{\left(s (s+1) I_n - A(s+1) - B \, G\right)} \\ & = (s+1)^{m-n} \det{\left(s^2 \, I_n + (I_n-A)s - A - B \, G\right)} \end{align*} $

It is kinda weird. My supervisor says the final result is awkward. xD

Answer 1

First of all, consider the expression

$$\det{(s^2I_n + A_1 s + A_0)} = 0.$$

One can observe that this is the characteristic polynomial of the matrix

$$ \begin{bmatrix}0 & I\\-A_0 & -A_1\end{bmatrix}, $$ which means that the eigenvalues of this matrix coincides with the roots of the characteristic polynomial above.

Otherwise, your calculations for the characteristic polynomial of $S$ is correct. I am not sure why your supervisor says that the result is "awkward" (which is meaningless in mathematics). I would suggest you to ask your supervisor why they think this result is awkward to get more insights.

Answer 2

This isn't a complete answer, but a possible direction to pursue.

In general for $n\times n$ matrices $A_1,\ldots,A_p$ and scalars (or indeterminates) $\alpha_1,\ldots,\alpha_p$, $$\bigwedge^m\bigl(\,\sum_{i=1}^p\alpha_i A_i\bigr)=\sum_r\alpha_1^{r_1}\cdots\alpha_p^{r_p}\,D_r(A_1,\ldots,A_p)\tag{1}$$ where the sum on the right is taken over all nonnegative partitions $r:r_1+\cdots+r_p=m$ and $D_r(A_1,\ldots,A_p)$ is the "mixed derivation" defined by $$D_r(A_1,\ldots,A_p)(x_1\wedge\cdots\wedge x_m)=\sum_sx_1\wedge\cdots\wedge A_ix_{s_i(1)}\wedge\cdots\wedge A_ix_{s_i(r_i)}\wedge\cdots\wedge x_m\tag{2}$$ where the sum on the right here is taken over all sequences $s_1,\ldots,s_p$ compatible with the partition $r$ (that is, $s_i$ is strictly increasing of length $r_i$ with values in $1,\ldots,m$ and the images of the $s_i$ are disjoint), and $A_i$ is in the positions indicated by $s_i$.

Taking the trace on both sides of (1) with $m=n$, we obtain $$\det\bigl(\,\sum_{i=1}^p\alpha_i A_i\bigr)=\sum_r\alpha_1^{r_1}\cdots\alpha_p^{r_p}\,\mathop{\mathrm{tr}}D_r(A_1,\ldots,A_p)\tag{3}$$

The trace on the right is the sum of the eigenvalues of $D_r(A_1,\ldots,A_p)$. The eigenvalues of $D_r(A_1,\ldots,A_p)$ are known to be partition polynomials of $r$ in the eigenvalues of $A_1,\ldots,A_p$.

In your case you can take $p=3$, $\alpha_i=s^{i-1}$, and $A_3=I_n$ in (3) to study the desired determinant form. The computation of eigenvalues is slightly simplified since one of the matrices is the identity matrix.