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Suppose $A$ is a $3\times 3$ symmetric matrix such that $$\begin{bmatrix}x & y &1\end{bmatrix} A \begin{bmatrix}x \\y & \\1 \end{bmatrix}=xy-1$$

How to prove $A$ has full rank and the number of positive eigenvalues of $A$ is one?

My try:

$$\begin{bmatrix}x & y &1\end{bmatrix} \begin{bmatrix}a & b &c \\b & d & e \\c & e & f \end{bmatrix} \begin{bmatrix}x \\y \\1 \end{bmatrix}=xy-1$$ $$\Rightarrow \begin{bmatrix}x & y &1\end{bmatrix} \begin{bmatrix}ax+by+c \\bx+dy+e \\cx+ey+f \end{bmatrix}=xy-1$$

$$\Rightarrow ax^2+2bxy+2cx+dy^2+2ey+f=xy-1 $$ $$\Rightarrow a=c=d=e=0\; \text{and} \;b=1/2, f=-1$$

So the matrix becomes

$$\begin{bmatrix}0 & 1/2 &0 \\1/2 & 0 & 0 \\0 & 0 & -1 \end{bmatrix}$$

which has only one positive eigenvalue, namely, $\frac 12$. Also this matrix is invertible and, hence, full rank.

Is this way correct or anything I'm doing wrong? What is the general idea?

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    $\begingroup$ I think what you have done is perfect! $\endgroup$ – vidyarthi Aug 6 '17 at 11:51
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Your argument is perfectly correct.

In general, you can convert a quadratic form $$ Ax^2 + 2B xy + C y^2 + 2Dx + 2Ey + F $$ into a matrix $$ M = \pmatrix{ A & B & D\\ B & C & E \\ D & E & F } $$ and the determinant of the matrix and trace can both be computed from the quadratic form, from which you can work out things like how many positive/negative eigenvalues there are.

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  • $\begingroup$ Oh! i really forget the concept "quadratic form".From your answer I remember it. Thanks! $\endgroup$ – Chinnapparaj R Aug 6 '17 at 11:58
  • $\begingroup$ @Chinz: I agree with others that your argument is just fine. But I think it is worth commenting that there is value in remembering the general structure of the matrix (matching up with the quadratic form) in John Hughes' answer so that one can write the matrix rather quickly and directly without going through the multiplications in your original post. $\endgroup$ – Just_to_Answer Aug 14 '17 at 2:31
  • $\begingroup$ @just-to-answer: Ok. I understand.. $\endgroup$ – Chinnapparaj R Aug 14 '17 at 2:38

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