Suppose $\underline{X}$ is an $m$-dimensional vector following multivariate Normal distribution i.e. $\underline{X}$~$N_m(\underline{\mu},\Sigma)$ where $\Sigma$ is positive definite. Let $B$ be a symmetric $m$ by $m$ matrix, having the following properties:
$B\Sigma B=B$
$\underline{\mu}'B\underline{\mu}=0$
$trace(B\Sigma)=k$, where $k<m$
Prove that $\underline{X}'B\underline{X}$ follows chi-squared ($k$).
I proceeded a lot but got stuck in one last step where I think linear algebra is troubling me. Kindly have a look at my "solution".
The second condition can be written as $\underline{\mu}'B\Sigma B\underline{\mu}=0\implies B\underline{\mu}=\underline{0}$ as $\Sigma$ is Positive definite. Then after a little algebra and considering condition 2, we can write $\underline{X}'B\underline{X}=(\underline{X}-\underline{\mu})'B(\underline{X}-\underline{\mu})=(\underline{X}-\underline{\mu})'B\Sigma B(\underline{X}-\underline{\mu})$
But $\Sigma$ is positive definite hence we can have a symmetric matrix $\sqrt{\Sigma}$ defined such that $\sqrt{\Sigma}\sqrt{\Sigma}=\Sigma$. Using this, we can write $(\underline{X}-\underline{\mu})'B\Sigma B(\underline{X}-\underline{\mu})=(\sqrt{\Sigma}B(\underline{X}-\underline{\mu}))'(\sqrt{\Sigma}B(\underline{X}-\underline{\mu}))$
We want to show that this follows chi-squared ($k$).
Now, it is obvious that as $(\underline{X}-\underline{\mu})$~$N(\underline{0},\Sigma)$, $\sqrt{\Sigma}B(\underline{X}-\underline{\mu})$~$N(\underline{0},\sqrt{\Sigma}B\Sigma B\sqrt{\Sigma})=N(\underline{0},\sqrt{\Sigma}B\sqrt{\Sigma})$
Now, $trace(\sqrt{\Sigma}B\sqrt{\Sigma})=trace(B\Sigma)=k$ and $\sqrt{\Sigma}B\sqrt{\Sigma}$ is idempotent (verified). So its eigenvalues are only $0$ or $1$. Since its trace is $k$, there are $k$ eigenvalues with value $1$ and other $m-k$ eigenvalues are $0$.
Now here is the thing I want to prove: I want to show that$\sqrt{\Sigma}B\sqrt{\Sigma}$, which is the covariance matrix, is diagonal and has only $k$ $1$'s and remaining $m-k$ diagonal entries $0$. The reason is, then I will be able to say that $k$ components of $\sqrt{\Sigma}B(\underline{X}-\underline{\mu})$ are independent standard normals whose sum of squares is our quadratic form and thus it follows chi-squared ($k$).
But I really could not show that $\sqrt{\Sigma}B\sqrt{\Sigma}$ has the desired property i.e. it is diagonal with rank $k$. But this is crucial, it is the finale. If this does not happen, then some components will be correlated with each other and we cannot say that our quadratic form has chi-squared ($k$).
I would really appreciate if someone helped me to prove this final part.
