We have seen that the linear operator \(T\) on \(\RR^2\) defined by \(T(x_1, x_2) = (x_1, 0)\) is represented in the standard ordered basis by the matrix \[ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}. \] This operator satisfies \(T^2 = T\). Prove that if \(S\) is a linear operator on \(\RR^2\) such that \(S^2 = S\), then \(S = 0\), or \(S = I\), or there is an ordered basis \(\mathfrak{B}\) for \(\RR^2\) such that \([S]_{\mathfrak{B}} = A\) (above).
Solution
Assume \(S^2=S\). For every \(v\in\mathbb R^2\), \[ v=Sv+(v-Sv), \] where \(Sv\in\operatorname{range}S\) and \[ S(v-Sv)=Sv-S^2v=0, \] so \(v-Sv\in\ker S\). Also, if \(w\in\operatorname{range}S\cap\ker S\), write \(w=Su\); then \[ w=Su=S^2u=S(w)=0. \] Therefore \[ \mathbb R^2=\operatorname{range}S\oplus\ker S. \] If the range is \(0\), then \(S=0\). If the kernel is \(0\), then the direct-sum decomposition forces the range to be all of \(\mathbb R^2\), and \(S^2=S\) with invertible \(S\) gives \(S=I\).
Otherwise both range and kernel are one-dimensional. Choose nonzero \(u\in\operatorname{range}S\) and nonzero \(v\in\ker S\). Then \(Su=u\) and \(Sv=0\), and \(\{u,v\}\) is a basis. In this basis \[ [S]=\begin{bmatrix}1&0\\0&0\end{bmatrix}. \]