Symmetric matrices in quadratic forms

Question

I have solved the following problem and I was able to solve it, but could not fully understand why the hint listed in the description is true.

If $y$ has the standard multivariate normal distribution, $y \sim N(o,I_n)$, then $y^TAy$ has a $\chi^2(p)$ distribution if and only if $A = A^2$ and rank$(A) = p$. Hint: It is without loss of generality to assume $A$ symmetric

Why is it without loss of generality to assume $A$ is symmetric? Or, is it true that $A$ 'operates' symmetrically in quadratic forms?

Answer 1

The reason that you can assume that $A$ is symmetric is that only the symmetric part of $A$ contributes to $y^TAy$. Observe that if $M^T=-M$, then since $y^TMy$ is a scalar, we have $$y^TMy = (y^TMy)^T = y^TM^Ty = -y^TMy,$$ therefore $y^TMy=0$. Now let $S=\frac12(A+A^T)$ and $M=\frac12(A-A^T)$. The former is symmetric and the latter skew-symmetric, and $A=S+M$. However, $y^TMy=0$, therefore $y^TAy=y^TSy$.

Answer 2

They meant iff $B^2=B$ and $rank(B)=p$ where $B=\frac{A+A^\top}{2}$ is symmetric and $y^\top Ay=y^\top By$.

The solution is to diagonalize $B= P D P^\top$ (where, since $B$ is real symmetric, $D$ is real diagonal and $P$ is real orthonormal).

The pdf of $z=P^\top y$ is $\pi^{-n/2} e^{-\|P z\|^2}=\pi^{-n/2} e^{-\|z\|^2}$ thus $z\sim N(0,I_n)$.

It is easy to conclude from there since we are now looking at $z^\top D z=\sum_{j=1}^n D_{jj}z_j^2$.