Let $X$ be an $n\times k$ real matrix with $\text{rank}(X)=k$, and let $Y$ be the $n^2\times nk$ matrix defined by
$$Y=[K_n +I_{n^2}][X\otimes I_n]$$
where $K_n$ denotes the commutation matrix of order $n$. Then $\text{rank}(Y)=nk-\frac{k(k-1)}{2}$.
Proof. Let $USV'$ be an SVD decomposition of $X$. Then
$$[U\otimes U][S\otimes I_n][V\otimes U]'$$ is an SVD decomposition of $X\otimes I_n$. Using the properties of the commutation matrix we get
$$Y=[U\otimes U][K_n +I_{n^2}][S\otimes I_n][V \otimes U]'$$
and so the rank of $Y$ is the same as the rank of $[K_n +I_{n^2}][S\otimes I_n]$. Now $S \otimes I_n= \begin{bmatrix} D\\ 0_{n(n-k)\times nk} \end{bmatrix}$ where $D$ is a diagonal matrix having $nk$ positive diagonal entries. Therefore the rank of $Y$ is the same as the rank of the matrix consisting of the first $nk$ columns of $K_n+I_{n^2}$.
Let $c_{ij}$ denote the $n(i-1)+j$ th column of $K_n+I_{n^2}$, where $1\leq i,j\leq n$. Using the explicit formula for the commutation matrix we have that
$$c_{i,j}=e_i \otimes e_j + e_j \otimes e_i$$
Since $e_i \otimes e_j=e_p \otimes e_q$ if and only if $(i,j)=(p,q)$ we see that $c_{i,j}=c_{p,q}$ if and only if $\{i,j\}=\{p,q\}$. Moreover if $c_{i,j}\neq c_{p,q}$ then $c_{i,j}$ and $c_{p,q}$ are orthogonal.
Therefore there are $nk-\frac{k(k-1)}{2}$ linearly independent columns among the first $nk$ columns of $K_n+I_{n^2}$.
Question: Which columns of $Y$ can be removed in order to obtain a full rank matrix?
Thanks a lot for your help.