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

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?

• I think what you have done is perfect! – vidyarthi Aug 6 '17 at 11:51

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.