# Linear Algebra Problem about Eigen-polynomial?

Prove that if $$Y= \begin{bmatrix} 0_n&A\\ A^T&0_{n+1}\\ \end{bmatrix}$$

For some $$n\times (n+1)$$ matrix $$A$$, define $$P(\lambda)=\det(Y-\lambda I), Q(\lambda)=\det(AA^T-\lambda I)$$. Prove that $$P(x)=x Q(x^2)$$.

I heard this problem is solvable with elementary matrix operations, but I can't get that to work. Instead, my approach seems very complicated:

Here is my approach: We first prove a lemma.

Lemma: Let $$n$$ be a positive integer, $$X=\begin{bmatrix} 0_n&A\\ B&0_n\\ \end{bmatrix}$$

Where $$A,B$$ are $$n\times n$$ matrices. Then $$\det(X)=(-1)^n \det(A)\det(B)$$

Let $$x_{i,j}$$ denote the entry on $$i$$th row and $$j$$th column of $$x$$. Define $$a_{i,j},b_{i,j}$$ similarly.

Let $$\sigma,\pi:[n]\to[n]$$ be permutations. Consider a permutation $$\tau:[2n]\to [2n]$$ such that $$\tau(j)=\sigma(j)+n$$ for $$1\le j\le n$$ and $$\tau(j+n)=\pi(j)$$. We can show $$sgn(\tau)=sgn(\sigma)sgn(\pi)(-1)^n$$ by transposing $$\tau,\sigma$$ or $$\tau,\pi$$ simultaneously.

Therefore, $$\sum_{\tau} sgn(\tau) \prod\limits_{j=1}^{2n} x_{j,\tau(j)} = (-1)^n \sum_{\sigma,\pi} sgn(\sigma)sgn(\pi) \prod\limits_{j=1}^{n} a_{j,\sigma(j)}\prod\limits_{k=1}^{n} b_{k,\pi(k)}$$

$$=(-1)^n \left(\sum_{\sigma} sgn(\sigma)\prod\limits_{j=1}^{n} a_{j,\sigma(j)} \right) \left(\sum_{\pi} sgn(\pi)\prod\limits_{k=1}^{n} b_{k,\pi(k)} \right)=(-1)^n \det(A)\det(B)$$

I can also prove via Hall's marriage theorem that if $$n\ne m$$, $$X= \begin{bmatrix} 0_n&A\\ B&0_m\\ \end{bmatrix}$$

Where $$A$$ is a $$n\times m$$ matrix, $$B$$ is a $$m\times n$$ matrix, then $$\det(X)=0$$.

Proof: I claim for every permutation $$\sigma:[n+m]\to [n+m]$$ there exists at least one $$1\le j\le n+m$$ such that $$x_{j,\sigma(j)}=0$$

Consider a bipartite graph. By Hall's marriage, done.

I think I ended up with a huge computational approach, but apparently this problem is doable with elementary row operations. I know what they are, but can you please tell me how to get elementary row operations to work?

• Hint: $(Y-xI)(Y+xI) = Y^2 - x^2I$ Aug 1, 2022 at 20:30

A sign is missing in the assertion. Let $$x$$ be an indeterminate. Since $$\pmatrix{I_n&\frac{1}{x}A\\ 0&I_{n+1}}\pmatrix{-xI_n&A\\ A^T&-xI_{n+1}} =\pmatrix{\frac{1}{x}AA^T-xI_n&0\\ A^T&-xI_{n+1}},$$ we have $$\det(Y-xI)=(-x)^{n+1}\det(\frac{1}{x}AA^T-xI_n)=(-1)^{n+1}x\det(AA^T-x^2I_n)$$.