Let \(M_{n\times n}\) be the vector space of real \(n\times n\) matrices. Prove that there are neighborhoods \(U\) and \(V\) of the identity matrix in \(M_{n\times n}\) such that for every \(A\in U\), there is a unique \(X\in V\) satisfying \[ X^4=A. \]
Solution
Identify \(M_{n\times n}\) with the real vector space \(\RR^{n^2}\), and define \[ \Phi:M_{n\times n}\to M_{n\times n}, \qquad \Phi(X)=X^4. \]
1. The map \(\Phi\) is continuously differentiable and satisfies\[ \Phi(I)=I. \]
Each entry of \(X^4\) is a polynomial in the entries of \(X\), so \(\Phi\) is a polynomial map and hence is continuously differentiable. The identity \(\Phi(I)=I\) is immediate.
2. The derivative of \(\Phi\) at \(I\) is\[ D\Phi_I(H)=4H. \]
For a matrix \(H\) and a real scalar \(t\), \[ \Phi(I+tH) =(I+tH)^4 = I+4tH+O(t^2) \] as \(t\to0\). Hence \[ \frac{\Phi(I+tH)-\Phi(I)}{t} \longrightarrow 4H, \] which gives the displayed derivative.
3. The linear map\[ D\Phi_I:M_{n\times n}\to M_{n\times n} \]is invertible.
By step 2, this map is multiplication by the nonzero scalar \(4\). Its inverse is multiplication by \(1/4\).
4. There are neighborhoods \(V\) and \(U\) of \(I\) such that\[ \Phi|_V:V\longrightarrow U \]is a bijection.
Steps 1 and 3 verify the hypotheses of the inverse function theorem at \(I\). Therefore there are neighborhoods \(V\) of \(I\) and \(U\) of \(\Phi(I)=I\) such that \(\Phi|_V\) is a diffeomorphism from \(V\) onto \(U\), in particular a bijection.
5. For every \(A\in U\), there is a unique \(X\in V\) satisfying\[ \boxed{X^4=A}. \]
By step 4, every \(A\in U\) has a unique preimage \(X\in V\) under \(\Phi\). By definition of \(\Phi\), the equation \(\Phi(X)=A\) is exactly \(X^4=A\).
6. Q.E.D.
Steps 4 and 5 give the required neighborhoods and the stated existence and uniqueness.