Difference between identity and a random projection

Question

Suppose a random projection $P$ in $\mathbb{R}^d$ onto a random n-dimensional 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 (d-n)-dimensional subspace uniformly distributed in Grassmannian $G_{d, d-n}$?

Answer 1

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 row-space of $X$) has $G_{n,d}$ distribution.

To see that $I_d-P$ has the $G_{d-n, n}$ 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 $X=\sum_{i=1}^n u_i s_i v_i^T$,
• $Y\in R^{(d-n)\times d}$ 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 row-space of $\tilde X = \sum_{i=1}^n u_i s_i b_i^T$, and $X=^d\tilde X$ (equality in distribution).
• $I_d-P= \sum_{i=n+1}^d b_i b_i^T$ is the row-space of $\tilde Y=\sum_{i=n+1}^d f_i \mu_i b_i^T$ and $Y=^d\tilde Y$.