Deterministic SVD-partition error bound for the range finder

Choose the SVD from [F1] and set $\tilde{\mathrm{\Omega }}=V^{\ast }\mathrm{\Omega }=[{\mathrm{\Omega }}_1^T{\mathrm{\Omega }}_2^T{]}^T$. Left multiplication by $U^{\ast }$ carries the sample range to $\mathrm{Ran}(\mathrm{\Sigma }\tilde{\mathrm{\Omega }})$, preserves the operator norm because it maps unit vectors bijectively to unit vectors by [F3], and carries $Q{Q}^{\ast }$ to the orthogonal projector $P$ onto that range. It suffices to prove the estimate for $\mathrm{\Sigma }$.

Since ${\mathrm{\Omega }}_1$ has full row rank, put ${\mathrm{\Omega }}_1^{†}={\mathrm{\Omega }}_1^{\ast }({\mathrm{\Omega }}_1{\mathrm{\Omega }}_1^{\ast }{)}^{-1}$, so ${\mathrm{\Omega }}_1{\mathrm{\Omega }}_1^{†}=I_k$. Let $Y=\mathrm{\Sigma }[{\mathrm{\Omega }}_1^T{\mathrm{\Omega }}_2^T{]}^T$ and $C=Y{\mathrm{\Omega }}_1^{†}[I_{k}0]$. The range of $C$ lies in $\mathrm{Ran}(Y)$, hence $(I-P)C=0$.

The first $k$ rows of $\mathrm{\Sigma }-C$ vanish because ${\mathrm{\Omega }}_1{\mathrm{\Omega }}_1^{†}=I_k$, and the remaining rows form $[-{\mathrm{\Sigma }}_2{\mathrm{\Omega }}_2{\mathrm{\Omega }}_1^{†}{\mathrm{\Sigma }}_2]$. Thus $\mathrm{\parallel }(I-P)\mathrm{\Sigma }{\mathrm{\parallel }}_2=\mathrm{\parallel }(I-P)(\mathrm{\Sigma }-C){\mathrm{\parallel }}_2\le \mathrm{\parallel }\mathrm{\Sigma }-C{\mathrm{\parallel }}_2\le \mathrm{\parallel }{\mathrm{\Sigma }}_2{\mathrm{\Omega }}_2{\mathrm{\Omega }}_1^{†}{\mathrm{\parallel }}_2+\mathrm{\parallel }{\mathrm{\Sigma }}_2{\mathrm{\parallel }}_2$: the horizontal block is a sum of two zero-padded blocks, so the triangle inequality applies. By [F1] and [F4], $\mathrm{\parallel }{\mathrm{\Sigma }}_2{\mathrm{\parallel }}_2={\sigma }_{k+1}$, including zero-tail and rank-deficient cases.