We are now in a position to fulfill our earlier promise to investigate the projections associated with the particular direct sum decompositions \(\mathcal{V}=\mathcal{M} \oplus \mathcal{M}^{\perp}\) . We shall call such a projection a perpendicular projection . Since \(\mathcal{M}^{\perp}\) is uniquely determined by the subspace \(\mathcal{M}\) , we need not specify both the direct summands associated with a projection if we already know that it is perpendicular. We shall call the (perpendicular) projection \(E\) on \(\mathcal{M}\) along \(\mathcal{M}^{\perp}\) simply the projection on \(\mathcal{M}\) and we shall write \(E=P_{\mathcal{M}}\) .
Theorem 1. A linear transformation \(E\) is a perpendicular projection if and only if \(E=E^{2}=E^{*}\) . Perpendicular projections are positive linear transformations and have the property that \(\|E x\| \leq \|x\|\) for all \(x\) .
Proof. If \(E\) is a perpendicular projection, then Section: Adjoints of projections , Theorem 1 and the theorem of Section: Dual of a direct sum show (after, of course, the usual replacements, such as \(\mathcal{M}^{\perp}\) for \(\mathcal{M}^{0}\) and \(A^{*}\) for \(A^{\prime}\) ) that \(E=E^{*}\) . Conversely if \(E=E^{2}=E^{*}\) , then the idempotence of \(E\) assures us that \(E\) is the projection on \(\mathcal{R}\) along \(\mathcal{N}\) , where, of course, \(\mathcal{R}=\mathcal{R}(E)\) and \(\mathcal{N}=\mathcal{N}(E)\) are the range and the null-space of \(E\) , respectively. Hence we need only show that \(\mathcal{R}\) and \(\mathcal{N}\) are orthogonal. For this purpose let \(x\) be any element of \(\mathcal{R}\) and \(y\) any element of \(\mathcal{N}\) ; the desired result follows from the relation \[(x, y)=(E x, y)=(x, E^{*} y)=(x, E y)=0.\] The positive character of an \(E\) satisfying \(E=E^{2}=E^{*}\) follows from \begin{align} (E x, x) &= (E^{2} x, x)\\ &= (E x, E^{*} x)\\ &= (E x, E x)\\ &= \|E x\|^{2}\\ &\geq 0. \end{align} Applying this result to the perpendicular projection \(1-E\) , we see that \begin{align} \|x\|^{2}-\|E x\|^{2} &= (x, x)-(E x, x)\\ &= \big([1-E] x, x\big)\\ &\geq 0; \end{align} this concludes the proof of the theorem. ◻
For some of the generalizations of our theory it is useful to know that idempotence together with the last property mentioned in Theorem 1 is also characteristic of perpendicular projections.
Theorem 2. If a linear transformation \(E\) is such that \(E=E^{2}\) and \(\|E x\| \leq\|x\|\) for all \(x\) , then \(E=E^{*}\) .
Proof. We are to show that the range \(\mathcal{R}\) and the null-space \(\mathcal{N}\) of \(E\) are orthogonal. If \(x\) is in \(\mathcal{N}^{\perp}\) , then \(y=E x-x\) is in \(\mathcal{N}\) , since \[E y=E^{2} x-E x=E x-E x=0.\] Hence \(E x=x+y\) with \((x, y)=0\) , so that \[\|x\|^{2} \geq\|E x\|^{2}=\|x\|^{2}+\|y\|^{2} \geq\|x\|^{2},\] and therefore \(y=0\) . Consequently \(E x=x\) , so that \(x\) is in \(\mathcal{R}\) ; this proves that \(\mathcal{N}^{\perp} \subset \mathcal{R}\) . Conversely, if \(z\) is in \(\mathcal{R}\) , so that \(E z=z\) , we write \(z=x+y\) with \(x\) in \(\mathcal{N}^{\perp}\) and \(y\) in \(\mathcal{N}\) . Then \[z=E z=E x+E y=E x=x.\] (The reason for the last equality is that \(x\) is in \(\mathcal{N}^{\perp}\) and therefore in \(\mathcal{R}\) .) Hence \(z\) is in \(\mathcal{N}^{\perp}\) , so that \(\mathcal{R} \subset \mathcal{N}^{\perp}\) , and therefore \(\mathcal{R} = \mathcal{N}^{\perp}\) . ◻
We shall need also the fact that the theorem of Section: Combinations of projections remains true if the word "projection" is qualified throughout by "perpendicular." This is an immediate consequence of the preceding characterization of perpendicular projections and of the fact that sums and differences of self-adjoint transformations are self-adjoint, whereas the product of two self-adjoint transformations is self-adjoint if and only if they commute. By our present geometric methods it is also quite easy to generalize the part of the theorem dealing with sums from two summands to any finite number. The generalization is most conveniently stated in terms of the concept of orthogonality for projections; we shall say that two (perpendicular) projections \(E\) and \(F\) are orthogonal if \(E F=0\) . (Consideration of adjoints shows that this is equivalent to \(F E=0\) .) The following theorem shows that the geometric language is justified.
Theorem 3. Two perpendicular projections \(E=P_{\mathcal{M}}\) and \(F=P_{\mathcal{N}}\) are orthogonal if and only if the subspaces \(\mathcal{M}\) and \(\mathcal{N}\) (that is, the ranges of \(E\) and \(F\) ) are orthogonal.
Proof. If \(E F=0\) , and if \(x\) and \(y\) are in the ranges of \(E\) and \(F\) respectively, then \[(x, y)=(E x, F y)=(x, E^{*} F y)=(x, E F y)=0.\] If, conversely, \(\mathcal{M}\) and \(\mathcal{N}\) are orthogonal (so that \(\mathcal{N} \subset \mathcal{M}^{\perp}\) ), then the fact that \(E x=0\) for \(x\) in \(\mathcal{M}^{\perp}\) implies that \(E F x=0\) for all \(x\) (since \(F x\) is in \(\mathcal{N}\) and consequently in \(\mathcal{M}^{\perp}\) ). ◻