The distance to the singular matrices in M_2(R)

Idea. $\|\cdot\|=|z|+|w|$ is a norm on $\mathbb{C}^2$ in which the singular set is $\{|z'|=|w'|\}$. The problem reduces to an elementary optimisation in the plane.

Step 1: the norm. By linearity of $A\mapsto(z,w)$ and $\|A\|=|z|+|w|$, if $B$ has coordinates $(z',w')$ then $\|A-B\|=|z-z'|+|w-w'|$.

Step 2: lower bound. Let $B$ be singular: $|z'|=|w'|=r$. Then, by the reverse triangle inequality $|z-z'|\ge||z|-r|$ and $|w-w'|\ge||w|-r|$, and by the triangle inequality in $\mathbb{R}$,
$$\|A-B\|\ge\bigl||z|-r\bigr|+\bigl||w|-r\bigr|\ge\bigl|(|z|-r)-(|w|-r)\bigr|=\bigl||z|-|w|\bigr|.$$

Step 3: the bound is attained. Assume $|z|\ge|w|$ with $(z,w)\ne(0,0)$ (the other case is symmetric by exchanging the roles, and $A=0$ is singular). Put $r=\frac{|z|+|w|}2$, $z'=r\,z/|z|$ (as $z\ne0$), and $w'=r\,u$ with $u=w/|w|$ if $w\ne0$ and $u=1$ if $w=0$. Then $|z'|=|w'|=r$ so the matrix $B$ with coordinates $(z',w')$ is singular, and
$$|z-z'|=|z|-r=\frac{|z|-|w|}2,\qquad|w-w'|=\bigl|\,|w|-r\bigr|=\frac{|z|-|w|}2.$$
(For $w=0$, $|w-w'|=r=\frac{|z|}2=\frac{|z|-|w|}2$.) Hence $\|A-B\|=|z|-|w|$.

Conclusion. The distance equals $\bigl||z|-|w|\bigr|$, which is $\sigma_2(A)$ by the previous problem.

Check (numerical). For random $A$, the matrix $B=A-\sigma_2u_2v_2^T$ (SVD) is singular and $\|A-B\|=\sigma_2$; moreover $300$ random rank-one matrices $uv^T$ never came closer than $\sigma_2$.

Remark. This is the $2\times2$ case of the Eckart-Young theorem: the distance from $A$ to the matrices of rank $<n$ is $\sigma_n(A)=1/\|A^{-1}\|$ for invertible $A$. In the $(z,w)$ picture, the nearest singular matrix is obtained by moving $z$ and $w$ along the radial lines to the common radius $r=\frac{|z|+|w|}2$.

Lesson. In the right coordinates, an "abstract" distance formula becomes a one-line triangle-inequality argument.