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?
