I came across the following exercise in preparation for a test.
Let $A \in \mathbb{R}^{n \times m}$ with rank $r < \min\{n,m\}$. Use SVD of $A$ to show that for every $\epsilon >0$ (no matter how small), there existsa full-rank matrix $A_{\epsilon} \in \mathbb{R}^{n\times m}$ such that $|| A - A_\epsilon ||_{2} < \epsilon$.
I am really puzzled on how to approach this exercise. From my understanding $$ A - A_\epsilon = \underset{n\times n}{U} \ \underset{n \times m}{\Sigma} \ \underset{m\times m}{V^{T}} - \underset{n\times n}{U} \ \underset{n \times m}{\Sigma_\epsilon} \ \underset{m\times m}{V^{T}} = U(\Sigma - \Sigma_\epsilon)V^{T}.$$ Then $\Sigma$ is diagonal matrix with singular values ranging from $\sigma_1$ to $\sigma_{\min\{n,m\}}$ and $\Sigma_\epsilon$ having singular values from $\sigma_1$ to $\sigma_m$ since $A_\epsilon$ is of rank $m$.
Then since $\min\{n,m\} \le m$ $$ \Sigma - \Sigma_\epsilon = \begin{pmatrix} \sigma_1 & & & \\ & \ddots & \\ & & \sigma_n & \\ & & & 0\end{pmatrix} - \begin{pmatrix} \sigma_1 & & & \\ & \ddots & \\ & & \sigma_{\min\{n,m\}} & \\ & & & 0\end{pmatrix} = \begin{pmatrix} 0& & & \\ & \ddots & \\ & & \sigma_? & \\ & & & 0\end{pmatrix}$$ So since a singular value exists for the matrix difference above, $|| A - A_\epsilon ||_2 = \sigma_?$ which has to be small since $\sigma_1 > \sigma_2 > \dots > \sigma_r$, and here this $\sigma_?$ is among among the last in the sequence.
I am most likely writing non-sense, I am aware of that, I don't know how to proceed, can anyone give me some hint or idea ?