Let \(M_{2\times2}(\mathbb R)\) be the four-dimensional vector space of real \(2\times2\) matrices and define \[ f:M_{2\times2}(\mathbb R)\to M_{2\times2}(\mathbb R), \qquad f(X)=X^2. \]
Show that \(f\) has a local inverse near \[ X=\begin{pmatrix}1&0\\0&1\end{pmatrix}. \]
Show that \(f\) does not have a local inverse near \[ X=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. \]
Solution
Write \[ I_2= \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}, \qquad J= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}. \]
1. For every \(X,H\in M_2(\RR)\),\[ Df_X(H)=XH+HX. \]
One has \[ \begin{aligned} f(X+H)-f(X) &= (X+H)^2-X^2\\ &= XH+HX+H^2. \end{aligned} \] For any matrix norm on the finite-dimensional space \(M_2(\RR)\), \[ \norm{H^2} \leq C\norm{H}^2 = o(\norm{H}) \] for some constant \(C\). Hence the linear part of the increment is \(H\mapsto XH+HX\).
2. Item 1: the derivative of \(f\) at \(I_2\) is invertible.
By step 1, \[ Df_{I_2}(H) = I_2H+HI_2 = 2H. \] Thus \(Df_{I_2}\) is scalar multiplication by \(2\) on the four-dimensional real vector space \(M_2(\RR)\), with inverse \(K\mapsto K/2\).
3. Item 1: the map \(f\) has a local inverse near \(I_2\).
The map \(f\) is polynomial and hence continuously differentiable. Step 2 shows that its derivative at \(I_2\) is an isomorphism. The inverse function theorem therefore gives neighborhoods \(U\) of \(I_2\) and \(V\) of \[ f(I_2)=I_2 \] such that \[ f|_U:U\longrightarrow V \] is a diffeomorphism. In particular, \(f\) has a local inverse near \(I_2\).
4. For every \(t\in\RR\), define\[ X_t= \begin{pmatrix} 1&t\\ 0&-1 \end{pmatrix}. \]Then\[ X_t^2=I_2. \]
Direct multiplication gives \[ X_t^2 = \begin{pmatrix} 1&t\\ 0&-1 \end{pmatrix} \begin{pmatrix} 1&t\\ 0&-1 \end{pmatrix} = \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} = I_2. \]
5. Item 2: the map \(f\) is not injective on any neighborhood of \(J\).
One has \[ X_0=J \] and \[ X_t\longrightarrow J \] as \(t\to0\). Therefore every neighborhood \(U\) of \(J\) contains some \(X_t\) with \(t\neq0\). For such \(t\), \[ X_t\neq J, \] but step 4 gives \[ f(X_t)=I_2=f(J). \] Hence \(f|_U\) is not injective.
6. Item 2: the map \(f\) does not have a local inverse near \(J\).
If a local inverse existed near \(J\), then \(f\) would be injective on some neighborhood of \(J\). Step 5 shows that this is impossible.
7. Therefore\[ \boxed{ f\text{ is locally invertible at }I_2 \text{ but not at }\operatorname{diag}(1,-1) }. \]
The first assertion is step 3 and the second is step 6.
8. Q.E.D.
Step 7 answers both items.