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?

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

1 Answer 1


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 $\det(Y-xI)=(-x)^{n+1}\det(\frac{1}{x}AA^T-xI_n)=(-1)^{n+1}x\det(AA^T-x^2I_n)$.

  • $\begingroup$ So is the centered equation similar to multiplication of a 2x2 matrix? Wow! I've never seen such manipulation before. $\endgroup$
    – Kai Wang
    Aug 1, 2022 at 23:28
  • $\begingroup$ @KaiWang Sort of. Tricks like this that make use of block matrix are sometimes quite handy. E.g. it is used in the derivation of Schur complement. $\endgroup$
    – user1551
    Aug 1, 2022 at 23:40
  • $\begingroup$ Yep. If two matrices are block-partitioned in a fashion for which matrix multiplication is legal, then the product can be written as a block matrix whose blocks are the products of the factors’ blocks. $\endgroup$
    – William
    Aug 2, 2022 at 3:02

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