Show that for a matrix $A$ there is a sequence of fully ranked matrices $\{ A_l \}_{l=1}^{\infty}$ such that $|| A-A_l||_2 \to 0$

Question

Let $A$ be a $m$ x $n$ matrix. I need to show that there exists a sequence of fully ranked matrices $\{ A_l \}_{l=1}^{\infty}$ such that $$|| A-A_l||_2 \to 0$$

I think I need to use here $SVD$ decomposition $A = U\Sigma V^*$, namely use the matrices $[A]_k$, when we take the first $k$ columns of $U$, the first $k$ singular values on $\Sigma$, and the firs $k$ rows of $V^*$.

The problem here is that the sequence is finite, so I am not really sure how that works. Maybe I just can denote $A_l = [A]_r$ for every $l \ge r$, when $r = rank(A)$.

Help would be appreciated.

Answer 1

Practically any random modification of $A$ has full rank, you do not need to use complicated tools like the SVD. However, if you want to go that way, use $$ A_l=U(Σ+I/l)V^∗, $$ assuming the SVD is of the economic type, $Σ$ square of size $\min(m,n)$.