How to find out number of positive eigenvalues of a symmetric matrix? Suppose $A$ is a $3 \times 3$ symmetric matrix such that $$[x,y,1]A\left[\begin{array}{c}
 x   \\
 y   \\
 1
\end{array}\right]=xy-1.$$
Let $p$ be the number of positive eigenvalues of $A$ and let $q = rank (A) - p$. Then 
(1) $p=1.$ 
2) $p=2.$ 
(3) $q=2.$ 
(4) $q=1.$
Since it is symmetric matrix the matrix is diagonalizable. Hence rank of A is 3.

We have $$A=\begin{bmatrix} 0 & \frac12 & 0 \\ \frac12 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$
Clearly $-1$ is an eigenvalue. The other eigenvalues satisfiesy $\lambda_1+\lambda_2=0$ and $\lambda_1\lambda_2=-\frac14$.  Hence the remaining eigenvalues are $\frac12$ and $-\frac12$.
$p=1$ and $q=3-1=2$.
