I've been playing around numerically with Haar random $\text{CUE}$ unitary matrices of size $N$ by $N$, with $N$ around $1000$. If I "truncate" the matrix by keeping the upper left $fN$ by $fN$ block for some fixed $f$ independent of $N$, setting all other entries to zero, the resulting matrix is no longer unitary. In fact, one can argue that the eigenvalues of the truncated matrix will with high probability now be strictly within, rather than on, the unit circle. The distribution of the eigenvalues is [well-known](https://arxiv.org/abs/0911.5645).


However, from my numerics, it appears that despite the truncation, $(2f-1)N$ singular values are still **exactly** $1$ for $f \geq 1/2$. When $f\leq 1/2$, all the singular values are less than one.


Now, the above seems true for generic unitaries, but there are fine-tuned exceptions. The simplest exception comes from taking the unitary $U$ to be the identity. Then there are exactly $fN>(2f-1)N$ unit singular values.


My question is, how can I see that the truncated $\text{CUE}$ matrices have, with high probability, $(2f-1)N$ singular values that are identically $1$?



The singular values of truncated Haar unitaries
Your numerical findings have a simple explanation: Let me denote the upper left block of the $N\times N$ unitary matrix by $R$ and the upper right block by $T$; the matrix $R$ has dimension $fN\times fN$, the upper right block has dimension $fN\times(1-f)N$; unitarity requires that
$$RR^\dagger+TT^\dagger=I,$$
with $\dagger$ indicating the conjugate transpose and $I$ the identity matrix. For $f\geq 1/2$ the matrix $TT^\dagger$ has rank $\leq (1-f)N$, so it has at least $fN-(1-f)N=(2f-1)N$ eigenvalues equal to zero.


Hence for $f\geq 1/2$ the matrix $RR^\dagger$ has at least $(2f-1)N$ eigenvalues equal to 1, or equivalently, $R$ has $\geq (2f-1)N$ singular values equal to unity. For generic matrices the $\geq$ will be an equality.


