Identify the real vector space \(M_{2\times2}(\mathbb R)\) with \(\mathbb R^4\) and define \[ F:M_{2\times2}(\mathbb R)\to M_{2\times2}(\mathbb R), \qquad F(X)=X+X^2. \] Prove that the range of \(F\) contains a neighborhood of the zero matrix.
Solution
Regard \(M_{2\times2}(\RR)\) as the real vector space \(\RR^4\).
1. The derivative of \(F\) at the zero matrix is the identity map on\(M_{2\times2}(\RR)\).
For \(H\in M_{2\times2}(\RR)\), \[ \begin{aligned} F(H)-F(0) &= H+H^2. \end{aligned} \] Since \[ \frac{\norm{H^2}}{\norm H} \leq \norm H \longrightarrow0 \qquad(H\to0) \] for any submultiplicative matrix norm, the linear part is \(H\). Hence \[ DF_0(H)=H. \] Thus \(DF_0\) is the identity and in particular is invertible.
2. There are neighborhoods \(U\) and \(V\) of the zero matrix such that\[ F:U\longrightarrow V \]is a diffeomorphism.
The map \(F\) is polynomial, hence continuously differentiable. By step 1, its derivative at \(0\) is invertible. The inverse function theorem therefore gives neighborhoods \(U\) of \(0\) and \(V\) of \[ F(0)=0 \] for which the displayed restriction is a diffeomorphism.
3. The range of \(F\) contains a neighborhood of the zero matrix.
By step 2, \[ V=F(U). \] Hence \[ V\subseteq F\bigl(M_{2\times2}(\RR)\bigr), \] and \(V\) is a neighborhood of \(0\).
4. Q.E.D.
Step 3 is the required conclusion.