Number of positive/negative/zero eigenvalues of a block matrix

Question

Let $B$ be an $n\times n$ matrix of rank $r$, and let $A$ be the $2n\times 2n$ matrix given by $$ A=\begin{bmatrix}I & B \\ B^t & 0 \end{bmatrix}.$$ How can we find the number of positive, negative, zero eigenvalues of $A$?

I have no idea for doing this, so I tried a computation with explicitly given $B$, and I am conjecturing that the answer will be as follows:

1. The number of the zero eigenvalues of $A$ equals that of $B$.

2. The number of positive eigenvalues of $A$ equals the number of negative eigenvalues of $B$ plus $n$.

3. The number of negative eigenvalues of $A$ equals the number of positive eigenvalues of $B$.

But I can't see how to prove this in the general situation. Any hints?

Answer 1

Sylvester's Law of Inertia

https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia

$$ \left( \begin{array}{cc} I & 0\\ -B^T&I \\ \end{array} \right) \left( \begin{array}{cc} I&B \\ B^T&0 \\ \end{array} \right) \left( \begin{array}{cc} I&-B \\ 0&I \\ \end{array} \right) = \left( \begin{array}{cc} I&0 \\ 0& - B^T B \\ \end{array} \right) $$

If $B$ is real, the rank of $B^T B$ is the rank of $B$ while the nonzero eigenvalues are positive. And then you negate it.