Let $p$ and $q$ be quadratic forms given by $$q(x,y,z,w)=x^2+y^2+z^2+bw^2$$ $$p(x,y,z,w)=x^2+y^2+czw$$ Then which of the following statements are true?
1) $p$ and $q$ are equivalent over $\mathbb{C}$, if $b$ and $c$ are non-zero complex numbers.
2) $p$ and $q$ are equivalent over $\mathbb{R}$, if $b$ and $c$ are nonzero real numbers.
3) $p$ and $q$ are equivalent non zero real numbers with $b$ negative.
4) $p$ and $q$ are not equivalent over $\mathbb{R}$, if $c$=0.
The corresponding matrices are $ Q= \left[ {\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & b \\ \end{array} } \right] $
$ P= \left[ {\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & \frac{c}{2} \\ 0 & 0 & \frac{c}{2} & 0 \\ \end{array} } \right]$
which are equivalent if and only if rank(P)=rank(Q) which is true when $b$ and $c$ are non-zero numbers. Therefore all the four options are correct. But the answer key says that option 2) is false. Can anyone help me find the mistake?
Hint- Let $p$ and $n$ respectively denote the number of positive and negative eigenvalues of a symmetric matrix $A$. Then $rank(A)=p+n$ and $signature (A)=p-n$.
For $\mathbb R$-equivalent, you need their rank equality and signature equality.