# What is the rank of a normally distributed matrix that is multiplied by rank-r projection matrices from left and right.

Let $$Z \in \mathbb{R}^{m_1\times m_2}$$ be a full rank matrix such that $$Z_{ij} \sim \mathcal{N}(0,1)$$. Moreover let $$U \in \mathbb{R}^{m_1\times R}$$ be a matrix with R orthonormal columns such that $$U^TU = I$$, and let $$V \in \mathbb{R}^{m_2\times R}$$ be a matrix with R orthonormal columns such that $$V^TV = I$$, where $$m_1,m_2 \geq R$$. Then, define two rank-R orthogonal projection matrices $$P_u = UU^T$$ and $$P_v = VV^T$$. The article I am reading states that it is easy to show that matrix $$X = P_uZP_v$$ has rank R almost surely. Unfortunately, I was unable to prove this statement and would appreciate any help. Thank you.

• The statement isn't necessarily true without some assumption on the multivariate distribution of the entries of $\ Z\$. Are they independent? Commented Jul 16, 2021 at 1:12
• Yes they are independent identically distributed. Commented Jul 17, 2021 at 4:02

Consider the case where $$m_1=m_2$$ is even and $$R=\frac{m_1}{2}$$. From here I select select $$Z:= I$$ and orthogonal matrix $$Q\in \mathbb R^{m_1\times m_1}$$

$$Q:=\bigg[\begin{array}{c|c|c|c|c} U &V\end{array}\bigg]$$

Then
$$X = P_uZP_v=\mathbf 0$$ and
$$R\gt \text{rank}\big(X)= 0$$
• How is $X$ equal to zero ? Commented Jul 14, 2021 at 1:14
• because $U^TV = \mathbf 0$ -- all of $Q$'s columns are orthonormal so any column in $U$ is orthogonal to any column in $V$ Commented Jul 14, 2021 at 4:22
• I forgot to specify that $Z_{ij} \sim \mathcal{N}(0,1)$, so probability of $Z$ being the identity matrix is $0$. Sorry. Commented Jul 15, 2021 at 23:55
• You are moving the goal post after the ball has been kicked, in fact 2 days after my answer. This is quite bad. Also, your new opening sentence "Let $Z \in \mathbb{R}^{m_1\times m_2}$ be a full rank matrix such that $Z_{ij} \sim \mathcal{N}(0,1)$" cannot be right. If $Z \in \mathbb{R}^{m_1\times m_2}$ then it has real scalars in each cell not random variables. I am done here. Commented Jul 16, 2021 at 0:01