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$\textbf{Question:}$

Prove that for every $A\in \mathbb{R}^{m\times n}$, $\exists$ a sequence of full rank matrices $A_i \in \mathbb{R}^{m\times n}$ such that $A_i \rightarrow A$ in $\|\cdot\|_2$.

$\textbf{Answer Given:}$

If $A$ is of full rank, then $A_i = A$ $\forall i$.

If A is not of full rank, then $A$ has the form

$$A= U \begin{bmatrix} \sigma_1 & 0 & \dots & \dots& 0\\ 0 & \sigma_2 & 0 &\dots & 0 \\ \vdots & & \ddots & & \vdots \\ 0 & \dots & 0 & \sigma_r & 0 \\ 0 & \dots & \dots & \dots & \mathcal{O} \end{bmatrix} V^T $$

where $\mathcal{O}$ is a $(m-r)\times (n-r)$ matrix of zeros.

Then for $m=n$,

$$A_i= U \begin{bmatrix} \sigma_1 & 0 & \dots & \dots& 0 & \dots & 0\\ 0 & \sigma_2 & 0 &\dots & 0 & & \vdots\\ \vdots & & \ddots & & \vdots & &\vdots\\ 0 & \dots & 0 & \sigma_r & 0 & & \vdots\\ 0 & \dots & \dots & \dots & \frac{1}{i} & & \vdots\\ \vdots & & & & & \ddots & \vdots\\ 0 & \dots & \dots & \dots & \dots & \dots & \frac{1}{i} \end{bmatrix} V^T $$

and for $m > n$,

$$A_i= U \begin{bmatrix} \sigma_1 & 0 & \dots & \dots& 0 & \dots & 0\\ 0 & \sigma_2 & 0 &\dots & 0 & & \vdots\\ \vdots & & \ddots & & \vdots & &\vdots\\ 0 & \dots & 0 & \sigma_r & 0 & & \vdots\\ 0 & \dots & \dots & \dots & \frac{1}{i} & & \vdots\\ \vdots & & & & & \ddots & \vdots\\ 0 & \dots & \dots & \dots & \dots & \dots & \frac{1}{i} \\ 0 & \dots & \dots & \dots & \dots & \dots & 0\\ \vdots & & & & & & \vdots\\ 0 & \dots &\dots & \dots & \dots & \dots & 0 \end{bmatrix} V^T $$

and for $m<n$,

$$A_i= U \begin{bmatrix} \sigma_1 & 0 & \dots & \dots& 0 & \dots & 0 & 0 & \dots & 0\\ 0 & \sigma_2 & 0 &\dots & 0 & & \vdots & \vdots & &\vdots\\ \vdots & & \ddots & & \vdots & &\vdots & \vdots & &\vdots\\ 0 & \dots & 0 & \sigma_r & 0 & & \vdots & \vdots & &\vdots\\ 0 & \dots & \dots & \dots & \frac{1}{i} & & \vdots & \vdots & &\vdots\\ \vdots & & & & & \ddots & \vdots & \vdots & &\vdots\\ 0 & \dots & \dots & \dots & \dots & \dots & \frac{1}{i} & 0 & \dots & 0 \end{bmatrix} V^T $$

Thus $\operatorname{rank}(A_i) = \min\{m,n\}$ and each $A_i$ is of full rank, which gives us

$$ \|A-A_i \|_2 = \begin{vmatrix} \begin{vmatrix} \begin{bmatrix} 0 & \dots & \dots & \dots & \dots & 0\\ \vdots & \ddots & & & & \vdots\\ 0 & & 0 \\ 0 & & 0 & \frac{1}{i} & \dots & 0 \\ \vdots & & & & \ddots & \vdots \\ 0 &\dots &\dots &\dots & 0 & \frac{1}{i} \end{bmatrix}\end{vmatrix}\end{vmatrix}_2 = \sigma_{i+1} = \frac{1}{i} \rightarrow 0 \text{ as } i \rightarrow \infty. $$

by the best rank-$k$ approximation theorem with respect to $\| \cdot \|_2$.

$\textbf{My Question (the only part that I do not understand):}$

How were the $\frac{1}{i}$ diagonals derived?

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They were by choice, with the goal of obtain $\|A-A_i\|_2=\frac1{i}$, in fact, you can replace $\frac1{i}$ by another positive function of $i$ that goes to $0$ as $i \to \infty$.

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  • $\begingroup$ Ah yes I apologize for the typos. I'll edit them right away! $\endgroup$ – Stoner Sep 2 '18 at 4:17
  • $\begingroup$ Also, is there a mathematical reason as to why we can simply pick any function of $i$ that goes to $0$ as $i \rightarrow \infty$? $\endgroup$ – Stoner Sep 2 '18 at 4:18
  • $\begingroup$ The goal is just to show the existence of a sequence of full rank matrices that goes to it right. It is like if the question is to ask show that there exists a sequence of positive sequence that goes to zero, we can either choose $\frac1{i}$ or $\frac1{i^2}$ and both should work. $\endgroup$ – Siong Thye Goh Sep 2 '18 at 4:22
  • $\begingroup$ Alright that makes sense to me now. Thanks again! :) $\endgroup$ – Stoner Sep 2 '18 at 4:39

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