For a real $2 \times 2$ matrix

$$
X=\left(\begin{array}{ll}
 x & y \\
 z & t
\end{array}\right),
$$

let $\|X\|=x^{2}+y^{2}+z^{2}+t^{2}$, and define a metric by $d(X, Y)=\|X-Y\|$. Let $\Sigma=\{X \mid \operatorname{det}(X)=0\}$. Let

$$
A=\left(\begin{array}{ll}
 1 & 0 \\
 0 & 2
\end{array}\right) \text {. }
$$

Find the minimum distance from $A$ to $\Sigma$.
For $X \in \Sigma$, we have

$$
\begin{aligned}
\|A-X\|^{2} &=(1-x)^{2}+y^{2}+z^{2}+(2-t)^{2} \\n&=y^{2}+z^{2}+1-2 x+x^{2}+(2-t)^{2} \\n& \geqslant \pm 2 y z+1-2 x+2|x|(2-t) \\n&=4|x|-2 x+2(\pm y z-|x| t)+1
\end{aligned}
$$

We can choose the sign, so $\pm y z-|x| t=0$ because $\operatorname{det} X=0$. As $4|x|-2 x \geqslant 0$, we have $\|A-X\| \geqslant 1$ with equality when $4|x|-2 x=0,|x|=2-t, y=\pm z$,

