Show that there exists \(\varepsilon>0\) such that, whenever \(A\) is a real \(2\times2\) matrix whose entries satisfy \[ \abs{a_{ij}}\le\varepsilon, \] there exists a real \(2\times2\) matrix \(X\) with \[ X^2+X^{\mathsf T}=A. \] Is \(X\) unique?
Solution
Let \(M=M_2(\RR)\), equip \(M\) with the Frobenius norm \(\norm{\cdot}_F\), and define \[ \Phi:M\longrightarrow M, \qquad \Phi(X)\coloneqq X^2+X^{\mathsf T}. \]
1. The derivative \(D\Phi_0:M\to M\) is the transpose map \(H\mapsto H^{\mathsf T}\), hence is a linear isomorphism.
For \(H\in M\), \[ \Phi(H)-\Phi(0)-H^{\mathsf T}=H^2. \] Since \(\norm{H^2}_F\leq\norm{H}_F^2\), the remainder is \(o(\norm{H}_F)\) as \(H\to0\). Thus \[ D\Phi_0(H)=H^{\mathsf T}. \] Transposition is its own inverse, so \(D\Phi_0\) is invertible.
2. There are open neighborhoods \(U,V\subset M\) of \(0\) such that \(\Phi:U\to V\) is a diffeomorphism.
The map \(\Phi\) is polynomial and therefore continuously differentiable. Step 1 shows that its derivative at \(0\) is invertible, so the inverse function theorem gives such neighborhoods \(U\) and \(V\).
3. There exists \(\varepsilon>0\) such that every real matrix \(A=(a_{ij})\) with \(\abs{a_{ij}}\leq\varepsilon\) lies in \(V\), and hence has a solution \(X\) of \(X^2+X^{\mathsf T}=A\).
Because \(V\) is an open neighborhood of \(0\), there is \(r>0\) such that the Frobenius ball \[ B_r(0)=\{A\in M:\norm{A}_F<r\} \] is contained in \(V\). Set \(\varepsilon=r/4\). If \(\abs{a_{ij}}\leq\varepsilon\) for all four entries, then \[ \norm{A}_F =\left(\sum_{i,j=1}^2 a_{ij}^2\right)^{1/2} \leq 2\varepsilon =\frac r2<r. \] Thus \(A\in V\). By step 2 there is a unique \(X\in U\) with \(\Phi(X)=A\).
4. The solution \(X\) is not unique in general.
Take \(A=0\). Besides \(X=0\), the matrix \[ Y=\begin{pmatrix}-1&0\\0&0\end{pmatrix} \] satisfies \[ Y^2+Y^{\mathsf T} =\begin{pmatrix}1&0\\0&0\end{pmatrix} +\begin{pmatrix}-1&0\\0&0\end{pmatrix} =0. \] Hence the equation can have more than one real solution, although step 3 gives a unique solution inside the neighborhood \(U\).
5. Q.E.D.
Step 3 proves the asserted local existence, and step 4 answers the uniqueness question.