The distribution of $\bf{x}$ given an underdetermined system $A{\bf x}={\bf b}\sim N(0,\sigma^2 I)$ Suppose I have an linear system $A{\bf x}={\bf b}$ such that $\bf b$ is a vector of IID normal random variables and $A$ has dimension $n\times p$ with $p > n$.
What can be said about the distribution of $\bf x$?  Must it come from a multivariate normal distribution?
Since the system is underdetermined one possible solution is certainly that $\bf x$ is multivariate normal, but since $A^TA$ is singular you can't explicitly solve for $\bf x$ as the sum of multivariate normal random variables.
We do have
$$A\mathbb{E}{\bf x}=0$$
$$A\text{Var}({\bf x})A^T=\sigma^2I$$
and thus the mean of $\bf x$ lies in the kernel of $A$.  Furthermore $\frac{1}{\sigma}A$ can be interpreted as the whitening matrix of $\bf x$, but that's about all I know.
 A: Consider a simple case where $A = \left[\begin{array}{cc}1& 1\end{array} \right] $ and $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.
Let $x_1$, $x_2$, and $b$ be random variables. We know from $A {\bf x} = {b}$ that $x_1 + x_2 =  b$. We also know that $b$ is a normally distributed random variable.


*

*Must $x_1$ and $x_2$ be distributed multivariate normal? No.

*Could $x_1$ and $x_2$ be multivariable normal? Yes.


In contrast, if $A \mathbf{x}$ is normal for any 1 by 2 matrix A, then $x_1$ and $x_2$ would be distributed multivariate normal (by definition of multivariate normality), but if $A$ is some specific matrix, all bets are off.

Example:
Let $b$ be a normally distributed random variable. Then define:
$${ x_1} = \left\{ \begin{array}{c} { b} &\text {if } {b} \leq 0 \\ 0 &\text{if } { b} > 0\end{array} \right\} $$
$${ x_2} = \left\{ \begin{array}{c} 0 &\text {if } { b} \leq 0 \\ { b} &\text{if } { b} > 0\end{array} \right\} $$
${x_1} + {x_2}$ is normally distributed but neither $x_1$ nor $x_2$ are normal.
