For \[ X=\begin{pmatrix}x&y\\z&t\end{pmatrix}, \] define \[ \|X\|=x^2+y^2+z^2+t^2, \qquad d(X,Y)=\|X-Y\|. \] Let \[ \Sigma=\{X:\det X=0\}, \qquad A=\begin{pmatrix}1&0\\0&2\end{pmatrix}. \] Find the minimum distance from \(A\) to \(\Sigma\), and exhibit a matrix \(S\in\Sigma\) attaining the minimum.
With \(\|X\|\) the sum of the squared entries, \(d(X,Y)=\norm{X-Y}_F^2\) is the square of the Frobenius distance and does not satisfy the triangle inequality. For the Frobenius distance \(\norm{X-Y}_F\) itself, the minimum distance from \(A\) to \(\Sigma\) is also \(1\), attained at the same \(S\).
Solution
For a real \(2\times2\) matrix \(B=(b_{ij})\), write \[ \norm{B}_F^2\coloneqq\sum_{i,j}b_{ij}^2. \] With the convention printed in the problem, \[ d(A,X)=\norm{A-X}_F^2. \]
1. For every real \(2\times2\) matrix \(B\) and every unit vector \(u\in\RR^2\),\[ \norm{Bu}_2^2\leq\norm{B}_F^2. \]
Let \(r_1,r_2\in\RR^2\) be the two rows of \(B\). By Cauchy–Schwarz, \[ \norm{Bu}_2^2 =\abs{r_1\cdot u}^2+\abs{r_2\cdot u}^2 \leq\norm{r_1}_2^2\norm{u}_2^2+\norm{r_2}_2^2\norm{u}_2^2. \] Since \(\norm u_2=1\), the right-hand side is \(\norm B_F^2\).
2. Every \(X\in\Sigma\) satisfies\[ d(A,X)\geq1. \]
Since \(X\) is singular, choose a unit vector \[ u=\begin{pmatrix}p\\q\end{pmatrix}\in\ker X. \] Apply step 1 to \(B=A-X\). Since \(Xu=0\), \[ d(A,X) =\norm{A-X}_F^2 \geq\norm{(A-X)u}_2^2 =\norm{Au}_2^2. \] But \[ \norm{Au}_2^2 =p^2+4q^2 \geq p^2+q^2 =1. \] Hence \(d(A,X)\geq1\).
3. The minimum distance is \(\boxed{1}\), attained by\[ \boxed{S=\begin{pmatrix}0&0\\0&2\end{pmatrix}}. \]
The matrix \(S\) is singular. Moreover, \[ A-S=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \] so the printed distance convention gives \[ d(A,S)=\norm{A-S}_F^2=1. \] Together with the lower bound in step 2, this proves that the minimum is \(1\) and that \(S\) attains it.
4. Q.E.D.
Step 3 gives both requested conclusions.