$X$ is a $p$-variate normal vector, and $A$ and $B$ are two symmetric matrices. When will $X^\prime AX$ and $X^\prime BX$ be independent? I have seen the answer $AB = 0$ is the condition, but I couldn't find an efficient proof.
 A: For efficiency, use characteristic functions.
Let $|\cdot|$ denote the determinant.  The following demonstration repeatedly evaluates the multinormal ($n$ dimensional) integral in the form
$$(2\pi)^{n/2}|T|^{n/2} = \int\cdots\int \exp\left(-\mathbf{x}\,T\,\mathbf{x}^\prime/2\right)\,\mathrm{d}^n\mathrm{x}$$
where $T$ is any symmetric matrix (with real or complex coefficients) for which the integral is finite.
The question of independence of functions of $X$ comes down to independence of the same functions of $X$ minus its mean (because the mean is constant), so we may assume the mean is zero.
Let's compute the characteristic functions of the variables $XAX^\prime$ and $XBX^\prime$ where the density of $X$ is proportional to $\exp(-\mathbf{x}V\mathbf{x}^\prime/2).$  (When the covariance $\Sigma$ is invertible, $V^{-1}=\Sigma.$  When $\Sigma$ is noninvertible all calculations are performed in a subspace where $\Sigma$ is supported and $A,B$ are understood as representing quadratic forms on this subspace.  Thus we may suppose $V$ is invertible.)
By definition, for any real number $s$ the characteristic function of $XAX^\prime$ is
$$\begin{aligned}
\phi_{XAX^\prime}(s) &=E\left[\exp(is\,XAX^\prime\right] \\
&=(2\pi|V|)^{-n/2}\int\cdots\int \exp\left(-\mathbf{x}\,\left(-2isA + V\right)\,\mathbf{x}^\prime/2\right)\,\mathrm{d}^n \mathrm{x} \\
&=|1_n - 2isAV^{-1}|^{n/2}.
\end{aligned}$$
Since random variables $(U,V)$ are independent if and only if $\phi_U(s)\phi_V(t) = \phi_{(U,V)}(s,t),$ the question of independence comes down to whether
$$|1_n - 2isAV^{-1}|^{n/2}\ |1_n - 2itBV^{-1}|^{n/2} = |1_n - 2isAV^{-1} - 2itBV^{-1}|^{n/2}.$$
Taking roots and using the multiplicative property of determinants shows this equation is equivalent to
$$|1_n - 2isAV^{-1} -2itBV^{-1} - 4stAV^{-1}BV^{-1}| =|1_n - 2isAV^{-1} - 2itBV^{-1}|$$
for all $(s,t).$ That clearly is the case if and only if the extra term on the left hand side vanishes; that is,

$$AV^{-1}\,BV^{-1} = 0.$$

(Equivalently, $A\Sigma B = AV^{-1}B = 0.$)
When $\Sigma=1_n$ is the identity matrix, that indeed is the condition $AB=0$ quoted in the question, QED.

BTW, notice that when $A$ is any real square matrix (symmetric or not), the quadratic form can be written
$$X A X^\prime = \frac{1}{2}X\left(A + A^\prime\right) X^\prime,$$
so we may replace $A$ by $(A+A^\prime)/2,$ which is symmetric, and proceed as before.  Therefore the requirement that $A$ and $B$ be symmetric is unnecessary.  The condition that the two forms be independent becomes
$$(A+A^\prime)\,\Sigma\,(B+B^\prime)= 0.$$
A: The result is called Craig's theorem. Named after the 1943 note Note on the Independence of Certain Quadratic Forms by Craig.
See this paper for details and history.
