Find smallest matrix E(as measured in the 2-norm), with the property that A-E is singular

Question

A: square nonsingular matrix with the SVD A=UΣV^T Find smallest matrix E(as measured in the 2-norm), with the property that A-E is singular.

Answer 1

Hint: Consider $E$ of the form $U \tilde{\Sigma} V^\top$ for some different diagonal matrix $\tilde{\Sigma}$. Among all such $E$ of this form that make $A-E$ singular, the smallest possible $2$-norm of $E$ will be $\sigma_{\min}(A)$. This will correspond to $\tilde{\Sigma}$ having all zero diagonal entries except a $\sigma_{\min}(A)$ in the same position as $\sigma_{\min}(A)$ in $A$.

Hint 2: It remains to show that if we consider arbitrary $E$, we still cannot find one with a smaller $2$-norm than the one explicitly constructed above. To do this, note that a Weyl inequality (see Exercise 22(iv) here) implies $$\sigma_{\min}(A) = |\sigma_{\min}(A-E) - \sigma_{\min}(A)| \le \|E\|_2.$$ (I am not certain if we need to appeal to Weyl's inequality to answer this question, so I am curious if there are simpler alternative approaches.)

Update: In Hint 2, I cited a rather strong result (Weyl inequality). But if we are looking at the minimum singular value (rather than the $i$th singular value), the inequality above can be proved directly, as shown here.