On problem regarding the quadratic form $yx+z^2$.

Question

If $A$ be a Matrix of rank $4$ and signature $3$, then there can not be an invertible Matrix $P$ such that $v^t(P^tAP)v$ is $xy+z^2$. Here $v \in R^4$.

Answer 1

The matrix B associated to $xy+z^2$ is

$$B=\begin{bmatrix}0&\frac12&0&0\\\frac12&0&0&0\\0&0&1&0\\0&0&0&0 \end{bmatrix}$$

with signature

$$n_+=2,n_-=1,n_0=1$$

and this is a contradiction since A and B should be congruent matrices.

Answer 2

Okay, by hypotheses, $A$ is congruent to $diag(1,1,1-1)$. [It has three positive eigenvalues, and no zero eigenvalues.]

By Sylvester's law of inertia, if $A$ were congruent to $$B=\begin{pmatrix} 0&1/2&0&0\\1/2&0&0&0\\0&0&1&0\\0&0&0&0\end{pmatrix}$$ then $B$ would have three positive eigenvalues and one negative one (and no zero eigenvalues). However, instead $B$ has two positive, one negative, and one zero eigenvalue. Hence $A$ and $B$ cannot be congruent.