Suppose $X \sim N_n(\mu, \sigma^2 I_n)$ and let $A$ and $B$ symmetric with shape $n\times n$, prove that $X^TAX$ is independent of $X^TBX$ if $AB=0$.
My attempt is that I have shown that $X^TAX$ is independent of $BX$ and $X^TBX$ is independent of $AX$. and it is not hard to prove that $AX$ is indep. of $BX$ by calculating covariance matrix on the basis that $AB=0$. But I got stuck that we can not show that $X^TAX$ indep. of $X^TBX$.
Besides that, I find a post may help. https://stats.stackexchange.com/questions/303466/prove-that-mathrmcovxtax-xtbx-2-mathrmtra-sigma-b-sigma-4-mu/476793#476793. It claims that $$ cov(X^TAX, X^TBX) = 4μTAΣBμ+2tr(AΣBΣ) $$ where it assume that $X \sim N(\mu, \Sigma)$. In my post, it reduces to be $$ cov(X^TAX, X^TBX) = 0 $$ But it is still unable to show that they are independent.
There exists an orthogonal matrix $U$ such that $U^TAU=\mathrm{diag}(a_1,\ldots,a_k,0,\ldots,0)$ where $A_1=\mathrm{diag}(a_1,\ldots,a_k)$ is invertible. Write $U^TBU=\left[\begin{array}{cc}B_1&C\\C^T&B_2\end{array}\right]$ by blocs with $B_1$ a $k\times k$ matrix. Then $$0=U^TABU=U^TAUU^TBU=\left[\begin{array}{cc}A_1B_1&A_1C\\0&0\end{array}\right]$$ which implies that $B_1$ and $C$ are zero since $A_1$ is invertible. Therefore $U^TBU=\left[\begin{array}{cc}0&0\\0&B_2\end{array}\right].$ Now write $X=m+\sigma Z$ where $Z\sim N(0,I_n)$ and
$$X=U\left[\begin{array}{c}X_1\\X_2\end{array}\right],\ m=U\left[\begin{array}{c}m_1\\m_2\end{array}\right],\ Z=U\left[\begin{array}{c}Z_1\\Z_2\end{array}\right]$$ The important point is that $Z_1$ and $Z_2$ are independent, as well as $X^TAX=m_1^TA_1m_1+2\sigma m_1^TA_1 Z_1+\sigma^2 Z_1^TA_1Z_1, \ X^TBX=m_2^TB_2m_2+2\sigma m_2^TB_2 Z_2+\sigma^2 Z_2^TB_2Z_2.$$