Exercise 15, § 1

Let $E$ be a complex Hilbert space, $E_1, E_2$ two closed vector subspaces of E, $P_1, P_2$ the orthogonal projectors of E onto $E_1, E_2$ respectively. (a) Now prove that, in order that $P_1$ and $P_2$ commute, it is necessary and sufficient that E be the Hilbert sum of the four subspaces $E_1 \cap E_2, E_1^\circ \cap E_2^\circ, E_1^\circ \cap E_2, E_1 \cap E_2^\circ$ (where $M^\circ$ denotes the orthogonal complement of a vector subspace M of E). (b) Now prove that if $E_1$ is of finite dimension and $\|P_1 - P_2\| < 1$, $E_2$ has the same dimension as $E_1$ (consider the intersection $E_1^\circ \cap E_2$). (c) Now prove that the endomorphism $T = (P_1 - P_2)^2$ of E commutes with $P_1$ and $P_2$, and that the proper subspace of $T$ corresponding to the eigenvalue 0 (resp. 1) is the direct sum of the orthogonal subspaces $E_1 \cap E_2$ and $E_1^\circ \cap E_2^\circ$ (resp. $E_1^\circ \cap E_2$ and $E_1 \cap E_2^\circ$). (d) Suppose that E is of finite dimension and that $T = \lambda I$ with $\lambda \neq 0$. Then $\lambda > 0$ and E is the Hilbert sum of subspaces of dimension $\leq 2$, each of which is stable by $P_1$ and $P_2$ (remark that $P_1 - P_2$ is hermitian and deduce that E is the Hilbert sum of two subspaces $E^+ , E^-$ such that $P_1 . x - P_2 . x = \sqrt{\lambda} x$ in $E^+$ and $P_1 . x - P_2 . x = -\sqrt{\lambda} x$ in $E^-$; now prove that for $x \in E^+$, one has $P_1 . x = \frac{1 + \sqrt{\lambda}}{2} x + z$ and $P_2 . x = \frac{1 - \sqrt{\lambda}}{2} x + z$, with $z \in E^-$). (e) Suppose that $E_1$ and $E_2$ are of finite dimension. Now prove that there exists a family $(F_\alpha)_{\alpha \in A}$ of subspaces of dimension $\leq 2$ of E such that E, $E_1$ and $E_2$ are respectively Hilbert sums of the families $(F_\alpha)_{\alpha \in A}$, $(F_\alpha \cap E_1)_{\alpha \in A}$ and $(F_\alpha \cap E_2)_{\alpha \in A}$ (use c) to reduce to the case where E is of finite dimension, then apply d)).