Consider the quadratic form $Q(v)=v^{t}Av,v=(x,y,z,w)$ where matrix $A$ is given by \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\\ \end{bmatrix}

Then which of the following is true?

1. $Q$ has rank 3.

2. $xy+z^{2}=Q(Pv)$ for some invertible real matrix $P.$

3. $xy+y^{2}+z^{2}=Q(Pv)$ for some real invertible matrix $P.$

4. $x^{2}+y^{2}-zw=Q(Pv)$ for some some real invertible matrix $P.$

It is clear that $Q$ has rank $4$ so $1$ is not true. How about other options. Please help me . Thanks in advance.

• I guess that $v = (x,y,z,w)$? – Friedrich Philipp Mar 4 '16 at 17:26
• yes yes ...i am going to edit..thanks – neelkanth Mar 4 '16 at 17:30

You have to find the matrices that generate the forms on the left hand sides. For example, $$xy + z^2 = \left\langle\left(\begin{matrix}0 & \frac 1 2 & 0 & 0\\\frac 1 2 & 0 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 0\end{matrix}\right)\left(\begin{matrix}x \\ y \\ z \\ w\end{matrix}\right),\left(\begin{matrix}x \\ y \\ z \\ w\end{matrix}\right)\right\rangle.$$ But this matrix has rank 3. Thus, no way.
• To be more precise: If $B$ is the found matrix above, you would have $\langle Bv,v\rangle = Q(Pv) = \langle APv,Pv\rangle = \langle P^TAPv,v\rangle$. By polarization, $B = P^TAP$ which cannot be true since $B$ has rank 3, but $P^TAP$ has rank 4 (since $P$ is assumed to be invertible). – Friedrich Philipp Mar 4 '16 at 17:42