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.

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 $1$?