The sum theorem for perpendicular projections is now easy.

Theorem 1. If \(E_{1}, \ldots, E_{n}\) are (perpendicular) projections, then a necessary and sufficient condition that \(E=E_{1}+\cdots+E_{n}\) be a (perpendicular) projection is that \(E_{i} E_{j}=0\) whenever \(i \neq j\) (that is, that the \(E_{i}\) be pairwise orthogonal).

Proof. The proof of the sufficiency of the condition is trivial; we prove explicitly its necessity only, so that we now assume that \(E\) is a perpendicular projection. If \(x\) belongs to the range of some \(E_{i}\) , then \begin{align} \|x\|^{2} &\geq \|E x\|^{2}\\ &=(E x, x)\\ &=\Big(\sum_{j} E_{j} x, x\Big) \\ &=\sum_{j}(E_{j} x, x)\\ &=\sum_{j}\|E_{j} x\|^{2}\\ &\geq\|E_{i} x\|^{2}\\ &=\|x\|^{2}, \end{align} so that we must have equality all along. Since, in particular, we must have \[\sum_{j}\|E_{j} x\|^{2}=\|E_{i} x\|^{2},\] it follows that \(E_{j} x=0\) whenever \(j \neq i\) . In other words, every \(x\) in the range of \(E_{i}\) is in the null-space (and, consequently, is orthogonal to the range) of every \(E_{j}\) with \(j \neq i\) ; using Section: Perpendicular projections , Theorem 3, we draw the desired conclusion. ◻

We end our discussion of projections with a brief study of order relations. It is tempting to write \(E \leq F\) , for two perpendicular projections \(E=P_{\mathcal{M}}\) and \(F=P_{\mathcal{N}}\) , whenever \(\mathcal{M} \subset \mathcal{N}\) . Earlier, however, we interpreted the sign \(\leq\) , when used in an expression involving linear transformations \(E\) and \(F\) (as in \(E \leq F\) ), to mean that \(F-E\) is a positive transformation. There are also other possible reasons for considering \(E\) to be smaller than \(F\) ; we might have \(\|E x\| \leq \|F x\|\) for all \(x\) , or \(F E=E F=E\) (see Section: Combinations of projections , (ii)). The situation is straightened out by the following theorem, which plays here a role similar to that of Section: Perpendicular projections , Theorem 3, that is, it establishes the coincidence of several seemingly different concepts concerning projections, some of which are defined algebraically while others refer to the underlying geometrical objects.

Theorem 2. For perpendicular projections \(E=P_{\mathcal{M}}\) and \(F=P_{\mathcal{N}}\) the following conditions are mutually equivalent. \begin{align} E &\leq F. \tag{i}\\ \|Ex\| &\leq \|Fx\| \text{ for all } x. \tag{ii}\\ \mathcal{M} &\subset \mathcal{N}. \tag{iii}\\ FE &= E, \tag{iv, a}\\ EF &= E. \tag{iv, b} \end{align} 

Proof. We shall prove the implication relations

(i) \(\implies\) (ii) \(\implies\) (iii) \(\implies\) (iv, a) \(\implies\) (iv, b) \(\implies\) (i).

(i) \(\implies\) (ii). If \(E \leq F\) , then, for all \(x\) , \[0 \leq \big([F-E] x, x\big)=(F x, x)-(E x, x)=\|F x\|^{2}-\|E x\|^{2}\] (since \(E\) and \(F\) are perpendicular projections).

(ii) \(\implies\) (iii). We assume that \(\|E x\| \leq \|F x\|\) for all \(x\) . Let us now take any \(x\) in \(\mathcal{M}\) ; then we have \[\|x\| \geq \|F x\| \geq \|E x\|=\|x\|,\] so that \(\|F x\|=\|x\|\) , or \((x, x)-(F x, x)=0\) , whence \[\big([1-F] x, x\big)=\|(1-F) x\|^{2}=0,\] and consequently \(x=F x\) . In other words, \(x\) in \(\mathcal{M}\) implies that \(x\) is in \(\mathcal{N}\) , as was to be proved.

(iii) \(\implies\) (iv, a). If \(\mathcal{M} \subset \mathcal{N}\) , then \(E x\) is in \(\mathcal{N}\) for all \(x\) , so that, \(F E x=E x\) for all \(x\) , as was to be proved.

That (iv, a) implies (iv, b), and is in fact equivalent to it, follows by taking adjoints.

(iv) \(\implies\) (i). If \(E F=F E=E\) , then, for all \(x\) , \[(F x, x)-(E x, x)=(F x, x)-(F E x, x)=\big(F[1-E] x, x\big)\] Since \(E\) and \(F\) are commutative projections, so also are \((1-E)\) and \(F\) , and consequently \(G=F(1-E)\) is a projection. Hence \[(F x, x)-(E x, x)=(G x, x)=\|G x\|^{2} \geq 0.\] This completes the proof of Theorem 2. ◻

In terms of the concepts introduced by now, it is possible to give a quite intuitive sounding formulation of the theorem of Section: Combinations of projections (in so far as it applies to perpendicular projections), as follows. For two perpendicular projections \(E\) and \(F\) , their sum, product, or difference is also a perpendicular projection if and only if \(F\) is respectively orthogonal to, commutative with, or greater than \(E\) .

EXERCISES

Exercise 1. 

  1. Give an example of a projection that is not a perpendicular projection.
  2. Give an example of two projections \(E\) and \(F\) (they cannot both be perpendicular) such that \(E F=0\) and \(F E \neq 0\) .

Exercise 2. Find the (perpendicular) projection of \((1,1,1)\) on the (one-dimensional) subspace of \(\mathbb{C}^3\) spanned by \((1,-1,1)\) . (In other words: find the image of the given vector under the projection onto the given subspace.)

Exercise 3. Find the matrices of all perpendicular projections on \(\mathbb{C}^{2}\) .

Exercise 4. If \(U=2 E-1\) , then a necessary and sufficient condition that \(U\) be an involutory isometry is that \(E\) be a perpendicular projection.

Exercise 5. A linear transformation \(U\) is called a partial isometry if there exists a subspace \(\mathcal{M}\) such that \(\|U x\|=\|x\|\) whenever \(x\) is in \(\mathcal{M}\) and \(U x=0\) whenever \(x\) is in \(\mathcal{M}^{\perp}\) .

  1. The adjoint of a partial isometry is a partial isometry.
  2. If \(U\) is a partial isometry and if \(\mathcal{M}\) is a subspace such that \(\|U x\|=\|x\|\) or \(0\) according as \(x\) is in \(\mathcal{M}\) or in \(\mathcal{M}^{\perp}\) , then \(U^{*} U\) is the perpendicular projection on \(\mathcal{M}\) .
  3. Each of the following four conditions is necessary and sufficient that a linear transformation \(U\) be a partial isometry.
    1. \(U U^{*} U=U\) ,
    2. \(U^{*} U\) is a projection,
    3. \(U^{*} U U^{*}=U^{*}\) ,
    4. \(U U^{*}\) is a projection.
  4. If \(\lambda\) is a proper value of a partial isometry, then \(|\lambda| \leq 1\) .
  5. Give an example of a partial isometry that has \(\frac{1}{2}\) as a proper value.

Exercise 6. Suppose that \(A\) is a linear transformation on, and \(\mathcal{M}\) is a subspace of, a finite-dimensional vector space \(\mathcal{V}\) . Prove that if \(\operatorname{dim} \mathcal{M} \leq \operatorname{dim} \mathcal{M}^{\perp}\) , then there exist linear transformations \(B\) and \(C\) on \(\mathcal{V}\) such that \(A x=(B C-C B) x\) for all \(x\) in \(\mathcal{M}\) . (Hint: let \(B\) be a partial isometry such that \(\|B x\|=\|x\|\) or \(0\) according as \(x\) is in \(\mathcal{M}\) or in \(\mathcal{M}^{\perp}\) and such that \(\mathcal{R}(B) \subset \mathcal{M}^{\perp}\) .)