Independence of a linear and a quadratic form How can I prove the following lemma?
Let $\mathbf{X}^ \prime$ = $ \left[ X_1 , X_2 , \ldots, X_n \right]$ where $ X_1, X_2, \ldots X_n $ are observations of a random sample from a distribution which is $N \left ( 0,\sigma^2 \right)$. Then let $\mathbf{b}^\prime = \left[b_1,b_2,\ldots,b_n \right]$ be a real nonzero vector and let $\mathbf{A}$ be a real symmetric matrix of order $n$. Then $\mathbf{b ^\prime X}$ and $ \mathbf{X} ^\prime \mathbf{A} \mathbf{X} $ are independent iff $\mathbf{b} ^\prime \mathbf{A}=0.$

I know that $\mathbf{b^ \prime X } \sim N(0, \sigma^2 \mathbf{ b^\prime b})$ but I do not see how I could proceed here. My main difficulty lies on the fact that these two variables are very different; had it been two quadratic forms instead, then Craig's theorem would be of use.
Any advice? 
Thank you.
 A: Use Craig's Theorem.  Consider the quadratic form on b.  If two random variables are independent, then any univariate functions of those random variables are likewise independent.  The quadratic forms are independent, ergo the linear form on b and the quadratic form on A are likewise independent.
A: Starting with the univariate case $X=X_1$, we find the correlation:
$\rho(bX,AX^2)=bA\rho(X,X^2)=bA\dfrac{\mathrm{Cov}(X,X^2)}{ \sigma_X \sigma_{X^2}} =bA\dfrac{E[(X-\mu_X)(X^2-\mu_{X^2})]}{ \sigma_X\sigma_{X^2}}$
With $\mu_x=0$, $\sigma_x=\sigma$, and for the expectation we know the distributions of $X$ and $X^2$ (Normal and Chi-Square).


*

*We see that $bA\neq 0$ implies $\rho\neq0$, so they are not independent in this case.

*Now look at the case $bA=0$, which means $b=A=0$, $(b=0,A\neq0)$ or $(b\neq0,A=0)$:
In the first case here,  we have $bX=0=AX^2$ and two constants are generally independent. For the other two cases, we have aswell that a constant $(0)$ and any random variable are independent.
So $(bX,AX^2)$ are independent only in the case $bA=0$, otherwise we have $\rho\neq0$.
This univariate case is not directly extendable to the multivariate case though because $X'AX\neq AX'X$.
As shortcut in general, if you have some transformation $Y=T(X)$, it is hence directly dependent on $X$ (not independent) except if any of them is a constant which here requires $b'A=0$.
