Let \(T\) be a linear operator on the finite-dimensional space \(V\). Suppose there is a linear operator \(U\) on \(V\) such that \(TU = I\). Prove that \(T\) is invertible and \(U = T^{-1}\). Give an example which shows that this is false when \(V\) is not finite-dimensional. (Hint: Let \(T = D\), the differentiation operator on the space of polynomial functions.)
Solution
Assume first that \(V\) is finite-dimensional and \(TU=I\). Then \(T\) is onto, because every \(v\in V\) equals \(T(Uv)\). For an endomorphism of a finite-dimensional space, surjectivity implies injectivity, hence \(T\) is invertible. Multiplying \(TU=I\) on the left by \(T^{-1}\) gives \[ U=T^{-1}. \]
The finite-dimensional hypothesis is essential. Let \(V=\mathbb R[x]\), let \(T=D\) be differentiation, and define \[ (Uf)(x)=\int_0^x f(t)\,dt. \] Then \(U\) maps polynomials to polynomials and \[ TU=D\circ U=I. \] But \[ UT(f)=\int_0^x f'(t)\,dt=f(x)-f(0), \] so \(UT\ne I\); for instance \(UT(1)=0\). Thus a right inverse need not be a left inverse in infinite dimension.