I am looking for an intuitive simple proof of the following lemma.
Suppose $x$ is an $n\times 1$ vector with iid $N(0,1)$ entries. Suppose $A$ is an $m\times n$ deterministic matrix with full row rank. Then, for any deterministic $b\in\mathbb R^m$, the conditional distribution $$x|_{Ax = b}\stackrel{d}{=} A'(AA')^{-1}b + P_{kernel(A)}(\tilde x)$$where $\tilde x$ has iid $N(0,1)$ components, and $P_S$ denotes projection onto subspace $S$.
I can derive this using the conditional Gaussian formula. But, I am looking for an alternative proof of this result which avoids computation. Here's a simpler way of looking at it but some details are incredulous to me.
Any vector $x$ satisfying $Ax=b$ can be decomposed as $x=x_0+n$ where $Ax_0=b$ and $n\in kernel(A)$. I understand that the first part of the right side in the desired equality is indeed this special $x_0$ but why this particular form? It turns out that this $x_0$ is the min-$\ell_2$ norm solution to $Ax=b$, and it's very surprising to me that this appears! This seems to be specific to normal? The second part of the decomposition is quite understandable given we understand where this first part comes from.
Again, I am not looking for any computation-driven answer (that is, directly computing the conditional distribution and performing algebra). I am rather looking for some insight into why this special solution $x_0$ comes up.
I don't know if this meets your requirements, but another way to view $x_0$ is to note that it is the projection of any $x$ satisfying $Ax = b$ onto the row-space of $A$.
$X$ can be uniquely expressed as $X_{0} + N$, where $X_0$ is the projection of $X$ onto the row space of $A$, and $N$ is the projection of $X$ onto the space orthogonal to the row space of $A$, which is precisely $\ker(A)$. Note that $X_0$ is determined by $AX$ and that $AX$ is uncorrelated with $N$. Now because $(AX,N)$ is normal, we can infer $N$ is infact independent of $AX$ and deduce the conditional distribution.