Combinations of projections

Continuing in the spirit of Theorem 3 of the preceding section, we investigate conditions under which various algebraic combinations of projections are themselves projections.

Theorem 1. We assume that and are projections on and along and respectively and that the underlying field of scalars is such that . We make three assertions.

is a projection if and only if ; if this condition is satisfied, then is the projection on along , where and .
is a projection if and only if ; if this condition is satisfied, then is the projection on along , where and .
If , then is the projection on along , where and .

Proof. We recall the notation. If and are subspaces, then is the subspace spanned by and ; writing implies that and are disjoint, and then ; and is the intersection of and .

If is a projection, then so that the cross-product terms must disappear: If we multiply (1) on both left and right by , we obtain subtracting, we get . Hence and are commutative, and (1) implies that their product is zero. (Here is where we need the assumption .) Since, conversely, clearly implies (1), we see that the condition is also sufficient to ensure that be a projection.

Let us suppose, from now on, that is a projection; by Section: Projections , Theorem 2, and are, respectively, the sets of all solutions of the equations and . Let us write , where and are in and , respectively, and and are in and , respectively. If is in , , then Since and , we have exhibited as a sum of a vector from and a vector from , so that . Conversely, if is a sum of a vector from and a vector from , then , so that is in , and consequently . Finally, if belongs to both and , so that , then so that and are disjoint; we have proved that .

It remains to find , that is, to find all solutions of . If is in , this equation is clearly satisfied; conversely implies (upon multiplication on the left by and respectively) that and . Since for all , we obtain finally , so that belongs to both and .

With the technique and the results obtained in this proof, the proofs of the remaining parts of the theorem are easy.

According to Section: Projections , Theorem 3, is a projection if and only if is a projection. According to (i) this happens (since, of course, is the projection on along ) if and only if and in this case is the projection on along . Since (2) is equivalent to , the proof of (ii) is complete.
That implies that is a projection is clear, since is idempotent. We assume, therefore, that and commute and we find and . If , then and similarly , so that is contained in both and . The converse is clear; if , then . Suppose next that ; it follows that belongs to , and, from the commutativity of and , that belongs to . This is more symmetry than we need; since , and since is in , we have exhibited as a sum of a vector from and a vector from . Conversely if is such a sum, then ; this concludes the proof that .

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We shall return to theorems of this type later, and we shall obtain, in certain cases, more precise results. Before leaving the subject, however, we call attention to a few minor peculiarities of the theorem of this section. We observe first that although in both (i) and (ii) one of and was a direct sum of the given subspaces, in (iii) we stated only that . Consideration of the possibility shows that this is unavoidable. Also: the condition of (iii) was asserted to be sufficient only; it is possible to construct projections and whose product is a projection, but for which and are different. Finally, it may be conjectured that it is possible to extend the result of (i), by induction, to more than two summands. Although this is true, it is surprisingly non-trivial; we shall prove it later in a special case of interest.

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