The distribution of the null space of a random Gaussian matrix Each element of a `fat' matrix is i.i.d standard normal distribution,  is the distribution of the element in its null space still normal?   For example,  $A$ is a $2\times 3$ matrix, each element of $A$ is normal distribution.  Obviously, there exists a $3\times 1$ vector $x$ that satisfies $Ax=0$.  So what is the distribution of the element in $x$?
 A: For the Gaussian case, the resulting vector $x$ can be represented as a Gaussian vector with correlated entries. 
Since $x$ satisfies the homogeneous equation $A x = 0$ for random Gaussian matrix $A$, it must equivalently satisfy the row-reduce form of $A$ as Gaussian elimination does not changes the solution space. 
Thus, assuming $A$ with dimensions of $m \times n, (m<n)$, then $x$ has $m$ independent variables (essentially, degrees of freedom) and  $n-m$ pivots that are just a linear combination of the $m$ independent Gaussian variables for any instance of $A$.  
Edit:
Following @antkam comment. Indeed, my original answer implicitly assumed that for any (given) Gaussian $A$ the resulting $x$ can be written as Gaussian as well. Specifically, following the complete probability formula, for any $x$ that satisfies $A x = 0$, we have, 
\begin{eqnarray*}
\Pr (x = \alpha) &=& \int_{A \in \mathcal{A}} f(A)\Pr (x = \alpha | A) \mathrm{d}A \\
&=& \int_{A \in \mathcal{A}} f(A)\delta_{\{A \alpha \}} \mathrm{d}A\\
&=& f(A \alpha)
\end{eqnarray*} 
where $f(A)$ is the matrix normal distribution, and $\delta(A \alpha)$ is Dirac's delta function.
Another interesting way to depict $x$ as multivariate Gaussian distribution is as follows. 
Let us consider the distribution of the vectors (basis) that spans the null-space of each row in the matrix $A$. Each such basis vector can be obtained by simply swapping two elements in a row vector of $A$, then, changing a sign of either one of the chosen elements, and finally, put zeroes in the remaining elements.  
For example, for matrix $$A = \begin{bmatrix}
    a_{11} & a_{12} & a_{13}  \\
    a_{21} & a_{22} & a_{23}  
\end{bmatrix}$$
The null space of the first row is spanned by the vectors $(a_{12}, - a_{11}, 0)$ and $(a_{13},0,-a_{11})$. The null space of the second row is spanned by the vectors $(a_{22}, - a_{21}, 0)$ and $(a_{23},0,-a_{21})$, and so on. 
Accordingly, each such basis follows the Gaussian distribution, since any linear combination of this vector's elements yield a Gaussian variable. Thus, the resulting null-space of each row in $A$ can be written as Gaussian. Note that other representations which, for example, involves row reduce operation will not necessarily yield Gaussian basis.  
Recall that the null-space that we seek is orthogonal to all the rows simultaneously. In other words, the null-space that we seek is the intersection of the null-spaces obtained from each row. Since each null-space can be written as Gaussian, the intersection can be written Gaussian as well.   
A: The space here is the distribution on $A_{m \times n}$, and the question is the distribution of $X$ s.t. $AX = 0$.  For this question to make sense, one first have to be able to determine $X$ given $A$, i.e. there must be a function (deterministic procedure) s.t. $X$ is uniquely determined by $A$ (i.e. a specific sample point).  Otherwise $X$ is not even a random variable and one cannot ask for its distribution.
If $m < n-1$, then the dimension of the null space of $A$ is at least 2 and IMHO it is not meaningful to pick a specific $X$ out of the null space.  
If $m = n-1$, then with probability 1 the dimension of the null space of $A$ is 1 and one can specify $X$ up to a scaling, so e.g. one can pick an $X$ s.t. $|X| = 1$ (there will be two of these, via scaling by -1) and one can reasonably ask what is the distribution on $X$.  This latter form of the question can be answered if the entries of $A$ are iid $N(0,1)$, as follows:
Consider a row of $A$.  If its entries are iid $N(0,1)$ then the joint pdf is only a function of the (Euclidean, i.e. 2-norm) length of the row, i.e., it is rotationally symmetric.  Thus the row specifies a random direction in $n$-space.  Any null space vector $X$ has to be perpendicular to this row.  There being $m=n-1$ rows, the null space vector $X$ has to be perpendicular to $n-1$ random directions, each chosen with rotational symmetry.  Thus $X$ itself is also rotationally symmetric.  Restricted to $|X|=1$ this means $X$ is uniform on the $(n-1)$ sphere.  In other words $X \sim Y/|Y|$ where $Y$ is a vector whose entries are iid $N(0,1)$.
If the entries have non-zero mean, then the rotational symmetry breaks down so I don't know what to say about $X$.
A completely different way to "interpret" this problem is to assume some prior distribution on $X$, e.g. its entries are iid $N(0,1)$, and then to ask what is the distribution of $X$ conditioned on $AX=0.$  The original wording did not suggest this, but this is one possible "interpretation" that made the original question meaningful.  Under this interpretation, and assuming all entries of $A$ are iid $N(0,1)$ (s.t. the distribution of its 1-dimensional null space is rotationally symmetric), and all entries of $X$ are iid $N(0,1)$ (s.t. its prior distribution is rotationally symmetric), then the rotational symmetry is preserved and the distribution of $X$ conditioned on $AX=0$ is unchanged.  However, again if entries of $A$ have non-zero mean then conditioned on $AX=0$ the distribution of $X$ will no longer have rotational symmetry (even if $X$ had rotational symmetry before conditioning).
