If $A$ is non-singular then there exist $B$ singular close to it

Question

The exercise says:

Show that if $A$ is non-singular, then there exist a singular matrix in a neighborhood with radio $\|A^{-1}\|^{-1}$ and center in $A$.

Can you help me with some hint for a way to prove it?

Answer 1

This is not true. If $A=I$ then every matrix $B$ with $\|B-A|| <1$ is non-singular.

Answer 2

Take the singular value decomposition of $A = U\Sigma V^T$, where if $\Sigma$ is $n\times n$ then $$ \Sigma = \mathrm{diag}(\sigma_1,...,\sigma_n) $$ and without loss of generality we take $\sigma_n = \min_i \sigma_i$. Then $$ \|A^{-1}\|^{-1} = \left(\frac{1}{\sigma_{n}}\right)^{-1} = \sigma_{n}>0. $$ This is exactly how far $A$ is from $B = U\hat \Sigma V^T$ where $$ \hat\Sigma = \mathrm{diag}(\sigma_1,\sigma_2,...,\sigma_{n-1},0). $$ and $B$ is singular.

Specifically, $\|A-B\|_2 = \sigma_n \underbrace{\|u_nv_n^T\|_2}_{=1} = \sigma_n$.