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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.

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

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  • $\begingroup$ okk sir now i will try ...thanks a lot.... $\endgroup$ – neelkanth Mar 4 '16 at 17:36
  • $\begingroup$ I edited. The matrix I had before was not symmetric. Now, it's ok. $\endgroup$ – Friedrich Philipp Mar 4 '16 at 17:38
  • $\begingroup$ yes now its ok.... $\endgroup$ – neelkanth Mar 4 '16 at 17:38
  • $\begingroup$ yes now i got the answer that is 4th one....thanks... $\endgroup$ – neelkanth Mar 4 '16 at 17:40
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    $\begingroup$ 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). $\endgroup$ – Friedrich Philipp Mar 4 '16 at 17:42

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