Suppose a random projection $P$ in $\mathbb{R}^d$ onto a random ndimensional subspace in $\mathbb{R}^d$ uniformly distributed in the Grassmannian $G_{d, n}$ (the projection of the row space of a random matrix $X \in \mathbb{R}^{n \times d}$ where each entry is i.i.d sampled from $\mathcal{N}(0, 1)$). Is $I  P$ a random projection onto a random (dn)dimensional subspace uniformly distributed in Grassmannian $G_{d, dn}$?

1$\begingroup$ If one were to use an alternative definition of the probability distribution on the Grassmannians as the distribution invariant under the action of the orthogonal group, then the answer to the question would be yes, because orthogonal complementation (i.e., sending $P$ to $IP$) commutes with the group action. I hope that the probability distribution specified in the question agrees with the invariant one, but I'm embarrassed to admit that I don't know that. So I also hope that an expert will come along and clarify this. $\endgroup$– Andreas BlassMay 25, 2021 at 14:40

$\begingroup$ The action of the orthogonal group leaves the distribution invariant: A rotation $Q\in O(d)$ acts on $X$ with $Y=XQ$ and $Y$ also has iid $N(0,1)$ entries. The projection into the rowspace of $Y$ is $Y^T(YY^T)^{1}Y$ and has image $Q$ times the image of $P=X^T(XX^T)^{1}X$. $\endgroup$– jlewkMay 25, 2021 at 15:39
1 Answer
The question defines a random subspace with $G_{n,d}$ distribution is as follows:
If $X$ with iid $N(0,1)$ admits the SVD decomposition $X=\sum_{i=1}^n u_i s_i v_i^T$ with $s_i>0$ then the image of $P=\sum_i v_i v_i^T$ (i.e., the rowspace of $X$) has $G_{n,d}$ distribution.
To see that $I_dP$ has the $G_{dn, n}$ distribution according to this definition, first observe that the three random matrices $(u_1,...,u_n)$, $(s_1,...,s_n)$ and $(v_1,...,v_d)$ are independent.
Next consider on a rich enough probability space:
 a large matrix $A\in R^{d\times d}$ with iid entries and SVD $A =\sum_{i=1}^d a_i \lambda_i b_i^T$,
 $X\in R^{n\times d}$ with iid $N(0,1)$ entries with SVD $X=\sum_{i=1}^n u_i s_i v_i^T$,
 $Y\in R^{(dn)\times d}$ with iid $N(0,1)$ entries with SVD $Y=\sum_{i=n+1}^d f_i \mu_i g_i^T$,
and assume that $(A,X,Y)$ are independent.
By independence observed above, we can combine the elements of the SVD of $A$, $X$ and $Y$ to produce coupled matrices:
 $P=\sum_{i=1}^n b_i b_i^T$ is the rowspace of $\tilde X = \sum_{i=1}^n u_i s_i b_i^T$, and $X=^d\tilde X$ (equality in distribution).
 $I_dP= \sum_{i=n+1}^d b_i b_i^T$ is the rowspace of $\tilde Y=\sum_{i=n+1}^d f_i \mu_i b_i^T$ and $Y=^d\tilde Y$.