What is the variance of $\mathrm{Proj}_{\mathcal{R}(A^T)}(z)$ for $z \sim \mathcal{N}(0, I_m)$ and $A_{ij} \overset{\text{iid}}{\sim} \mathcal N (0,1/m)$?
Let $m \ll N$ and suppose that $A \in \mathbb R^{m\times N}$ is a matrix whose entries are [normalized] iid normal, $A_{ij} \sim \mathcal{N}(0, 1/m)$. Suppose that $z \in \mathbb R^m$ has iid standard normal entries $z_i \sim \mathcal N(0,1)$ and denote the $z$-shifted null space of $A$ by $$ N_z := \{ \zeta \in \mathbb R^N : A \zeta = z\}. $$ For a vector $w \in \mathbb R^m$, denote the orthogonal projection of $w$ onto the orthogonal complement of the null space of $A$ by $$ \mathcal P_{\mathcal R(A^T)}(w) := {\arg\min} \big\{ \|\eta \|_2 : A\eta = w \big\} $$
Given $\mathrm{Var}(z) = I_m$, $\mathrm{Var}(A_{ij}) = 1/m$, and any $\eta\in N_z$, what is $\mathrm{Var}(\mathrm{Proj}_{\mathcal{R}(A^T)}(z))$?
Fortunately, $\mathrm{Proj}_{\mathcal{R}(A^T)}(z) = A^T z$. By independence, $\mathbb E[ (A^T z)_j ] = \sum_{i=1}^m \mathbb E [ A_{ij} ] \mathbb E [ z_i] = 0$. Hence, relying on independence once more, $$ \mathrm{Var}((A^Tz)_j) = \mathbb E[(A^Tz)_j^2] = \sum_{i=1}^m \mathbb E [ A_{ij}^2 ] \mathbb E [ z_i^2 ] = 1. $$