No. of positive eigen values of $3\times 3$ real matrix Suppose $A$ is a $3\times 3$ symmetric matrix such that  :
$$\begin{pmatrix} a & b & 1 
 \end{pmatrix} A  \begin{pmatrix} a  \\ 
b\\
1\end{pmatrix}  = ab -1$$ for all $a,b\in \mathbb{R}$
Question is to find no. of positive eigen values of $A$. and rank of $A$.
What i have observed so far is  :


*

*As $A$ is symmetric matrix, it should have real eigen values. item

*As $x^TAx$ is not positive, for $x= \begin{pmatrix} a  \\ 
b\\
1\end{pmatrix}$, ($a=b=\frac{1}{2}$) we see that not  all eigen values of $A$ are positive

*As $x^TAx$ is not negative, for $x= \begin{pmatrix} a  \\ 
b\\
1\end{pmatrix}$, ($a=2,b=1$) we see that not  all eigen values of $A$ are Negative


So, $A$ does have positive eigen values and also negative eigen values.
Now, I have little less clarity particularly here :
If i take $a=b=1$ the, I have 
$$\begin{pmatrix} 1 & 1 & 1 
 \end{pmatrix} A  \begin{pmatrix} 1  \\ 
1\\
1\end{pmatrix}  = 1 -1=0$$
I want to conlcude from this that $$ A  \begin{pmatrix} 1  \\ 
1\\
1\end{pmatrix}  =0\begin{pmatrix} 1  \\ 
1\\
1\end{pmatrix}$$ somehow converting row marrix in one side to column matrix in other side.
If this is true, then i would have $0$ as one eigen value.
So, I have one positive, one negative, one $0$ eigen value.
So, Rank of $A$ should be $2$ and no. of positive eigen values is $1$.
Please help me to sort out some mistakes (if there are any) and help me to make this answer a little messier than what i have written.
This problem is already asked some time ago but, the answer for that was "to try with a  brute force" which i thought is not the only way to go.
I had some start up but that is not entire answer, so i can not post that as an answer there.. 
So, I thought i should ask here at the risk of getting negative votes.
P.S : Here is the link for a copy of this what is no. of positive eigen value of symmetric matrix A with some given relationship
 A: The condition
$$\begin{pmatrix} a & b & 1 
 \end{pmatrix} A  \begin{pmatrix} a  \\ 
b\\
1\end{pmatrix}  = ab -1$$
for all $a$ and $b$ implies that $A$ must have the form
$$A = \begin{pmatrix}0 & p & q \\ 1-p & 0 & r \\ -q & -r & -1 \end{pmatrix}$$
(consider the nine elements of $A$ as unknowns, and equate coefficients in the polynomials in $a$ and $b$ on each side of the equation).
There is only one set of values for $p$, $q$ and $r$ that make $A$ symmetric. Now that you know $A$ explicitly, you can compute its rank, eigenvalues, etc. directly.

If this is too simple for you, you can also multiply your equation by $z^2$ to find
$$\begin{pmatrix} za & zb & z 
 \end{pmatrix} A  \begin{pmatrix} za  \\ zb\\ z\end{pmatrix}  = abz^2 - z^2$$
and then for a nonzero $z$ and arbitrary $x$ and $y$ set $a=x/z$ and $b=y/z$ to find
$$\begin{pmatrix} x & y & z 
 \end{pmatrix} A  \begin{pmatrix} x  \\ y \\ z\end{pmatrix}  = xy - z^2$$
which by continuity of each side must also hold for $z=0$. Now you have the quadratic form written down explicitly in full generality, and it is trivial to write it down as a symmetric matrix $A$. Again its eigenvalues and -vectors are then easy to find directly.
