in going through some books about numerical mathematics I found the following exercise:
Let $A,B \in \mathbb{R}^{n\times n}$ with $A$ symmetrical and rank($A$) = rank(B) = $n$. Define $M = \left[\begin{matrix} A & B \\ B^T & 0 \end{matrix}\right] $.
The statement now is, that $M$ has exactly $n$ positive and $n$ negative eigenvalues. And to prove it one should use the Courant minimax principle.
First we can see, that $M$ ist symmetrical, so Courant minimax principle is actually applicable for $M$. As the eigenvalues given by the minimax principle are ordered ($\lambda_1 \geq \lambda_2 \geq \dots \geq \lambda_{2n}$) we only have to consider two of them: $\lambda_n$ and $\lambda_{n+1}$. If $\lambda_n$ is positive and $\lambda_{n+1}$ negative, then the statement is proven right.
And this is where my problems begin. The Courant minimax principle gives that $\lambda_k = \max\limits_{U \in \mathcal{U}_k} \min\limits_{x \in U, ||x|| = 1}{x^T M x}$, where $\mathcal{U}_k = \{ U \subset \mathbb{R}^{2n} | \dim{U} = 2n + 1 -k\}$. I can't figure out how to make any kind of estimate for the value of $\lambda_k$ (in particular for $k = n$ and $k = n+1$) given how little we know about the matrix $M$ or it's components $A$, and $B$.
(Too long for a comment.) I don't know why we "should" apply the Courant-Fischer minimax principle. The statement can be proved pretty easily by Sylvester's law of inertia (which is a much weaker statement than the Courant-Fischer minimax principle).
Since $B$ has full rank, by matrix congruence (consider $\pmatrix{I&-\frac12A(B^T)^{-1}\\ 0&I}$), you may assume that $A=0$. In this case, the block matrix is clearly nonsingular. Now, if $(x^T,y^T)$ (with $x,y\in\mathbb R^n$) is an eigenvector for a nonzero eigenvalue $\lambda$, then $(x^T,-y^T)$ is an eigenvector for the eigenvalue $-\lambda$. So, eigenvalues of the block matrix occur in pairs of nonzero values with equal moduli but opposite signs. Hence the result.