How to be sure that the $k$th largest singular value is at least 1 of a matrix containing a k-by-k identity In section 8.4 of the report of ID software, it says that the $k$th largest singular value of a $k \times n$ matrix $P$ is at least 1 if some subset of its columns makes up a $k\times k$ identity.
I tried to figure it out but couldn't be sure of that. Any ideas on how to prove it?

The middle part of the SVD (which containts the singular values) does not change if you permute columns, so you may put the $k$ columns mentioned first. So assume the matrix has the form $A=[\begin{smallmatrix}I&B\end{smallmatrix}]$ where $I$ is a $k\times k$ identity, and $B$ is $k\times(n-k)$. Compute $$AA^T=I+BB^T\ge I$$
and conclude that eigenvalues of $AA^T$ are all at least $1$.
