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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

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marked as duplicate by hardmath, Peter Taylor, Daniel Fischer, Lord_Farin, Dan Rust Oct 12 '13 at 14:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The problem is that you can't simply go from $x^TAx=0$ to $Ax=0$. – egreg Oct 12 '13 at 13:18
@egreg : Yes, Yes.. I see that problem.. I have no other idea to go any further... – Praphulla Koushik Oct 12 '13 at 13:21
The context of "to try with brute force" in the previous Question's Answer was to use variables for the entries of $A$ and use the condition on the quadratic form $v^T A v$ to determine those entries. As Henning shows more explicitly, with symmetry of $A$, this approach succeeds. – hardmath Oct 12 '13 at 13:40

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.

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I don't want to compute explicitly what are the eigen values rather, I want to use more theory to conclude something about the given question.. Anyways Thanks for helping :) – Praphulla Koushik Oct 12 '13 at 13:26
@Praphulla: Knowing $A$ explicitly would at least prevent you from trying to prove falsehoods such as $A(1~1~1)^T=0$. – Henning Makholm Oct 12 '13 at 13:31
Yes, Yes... I totally agree with your view and thanks for clarifying that $(1 1 1)^T$ is not eigen vector corresponding to $0$ (if 0 is actually an eigen value) – Praphulla Koushik Oct 12 '13 at 13:35

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