On the block diagonalization of semi-unitary matrix

Suppose that we are given a $nk\times nm$ semi-unitary matrix $\mathbf{F}$ with $k \geq m$, i.e., $\mathbf{F}^H\mathbf{F}=\mathbf{I}_{nm}$, where $^H$ represents matrix conjugate transposition and $\mathbf{I}_{nm}$ is the $nm$-dimensional identity matrix. Is it possible to transform $\mathbf{F}$ into a block diagonal form, i.e., there exists $N$ unitary matrices $\mathbf{U}_i\in\mathbb{C}^{nm\times nm},i=1,\ldots,N$, such that $\mathbf{F}\mathbf{U}_{1}\mathbf{U}_2\cdots\mathbf{U}_N=\mathrm{blkdiag}\{\mathbf{F}_1,\mathbf{F}_2,\ldots,\mathbf{F}_n\}$ with $\mathbf{F}_i\in \mathbb{C}^{k\times m}, i=1,\ldots,n$.

• if $F$ is not a square matrix, how you define a block diagonal matrix? I assume that when you write $F={\rm blkdiag}\{F_1,\ldots,F_n\}$ you means instead $FU_1U_2\cdots U_N={\rm blkdiag}\{F_1,F_2,\ldots,F_n\}$, right? – Masacroso Dec 7 '17 at 2:51
• Yes. Thank you for the comment. I have revised the original problem formulation. Because $\mathbf{F}_i$ is of dimension $k\times m$ for $i=1,\ldots,n$, it makes sense to place them on the diagonal of $\mathbf{F}\mathbf{U}_1\mathbf{U}_2\cdots\mathbf{U}_N$. – Benjamin Mak Dec 7 '17 at 3:00
• If we place $\mathbf{F}_i\in\mathbb{C}^{k\times m}, i=1,\ldots,n$ on the diagonal of $\mathbf{F}\mathbf{U}_1\mathbf{U}_2\cdots\mathbf{U}_N\in\mathbb{C}^{nk\times nm}$, the dimensions add up. – Benjamin Mak Dec 7 '17 at 3:14
• The dimension of $\mathbf{F}$ is $nk\times nm$. Multiplied by $nm\times nm$ unitary matrices on the right, i.e., $\mathbf{F}\mathbf{U}_1\mathbf{U}_2\cdots\mathbf{U}_N$, the resulting matrix has a dimension of $nk\times nm$. – Benjamin Mak Dec 7 '17 at 3:30
• ah, you are right, sorry. – Masacroso Dec 7 '17 at 3:35