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.
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$\begingroup$ The statement isn't necessarily true without some assumption on the multivariate distribution of the entries of $\ Z\ $. Are they independent? $\endgroup$– lonza leggieraCommented Jul 16, 2021 at 1:12
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$\begingroup$ Yes they are independent identically distributed. $\endgroup$– karate_kidCommented Jul 17, 2021 at 4:02
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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
$R\gt \text{rank}\big(X)= 0$
which contradicts your claim.
answered Jul 13, 2021 at 0:37
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$\begingroup$ because $U^TV = \mathbf 0$ -- all of $Q$'s columns are orthonormal so any column in $U$ is orthogonal to any column in $V$ $\endgroup$ Commented Jul 14, 2021 at 4:22
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$\begingroup$ I forgot to specify that $Z_{ij} \sim \mathcal{N}(0,1)$, so probability of $Z$ being the identity matrix is $0$. Sorry. $\endgroup$ Commented Jul 15, 2021 at 23:55
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$\begingroup$ 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. $\endgroup$ Commented Jul 16, 2021 at 0:01
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$\begingroup$ I just forgot something and added it. I said I am sorry. No need to be angry. $\endgroup$ Commented Jul 16, 2021 at 0:24