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 E_{1} and E_{2} are projections on \mathcal{M}_{1} and \mathcal{M}_{2} along \mathcal{N}_{1} and \mathcal{N}_{2} respectively and that the underlying field of scalars is such that 1+1 \neq 0 . We make three assertions.
- E_{1}+E_{2} is a projection if and only if E_{1} E_{2}=E_{2} E_{1}=0 ; if this condition is satisfied, then E=E_{1}+E_{2} is the projection on \mathcal{M} along \mathcal{N} , where \mathcal{M}=\mathcal{M}_{1} \oplus \mathcal{M}_{2} and \mathcal{N}=\mathcal{N}_{1} \cap \mathcal{N}_{2} .
- E_{1}-E_{2} is a projection if and only if E_{1} E_{2}=E_{2} E_{1}=E_{2} ; if this condition is satisfied, then E=E_{1}-E_{2} is the projection on \mathcal{M} along \mathcal{N} , where \mathcal{M}=\mathcal{M}_{1} \cap \mathcal{N}_{2} and \mathcal{N}=\mathcal{N}_{1} \oplus \mathcal{M}_{2} .
- If E_1E_2 = E_2E_1 = E , then E is the projection on \mathcal{M} along \mathcal{N} , where \mathcal{M} = \mathcal{M}_1 \cap \mathcal{M}_2 and \mathcal{N} = \mathcal{N}_1 \cap \mathcal{N}_2 .
Proof. We recall the notation. If \mathcal{H} and \mathcal{K} are subspaces, then \mathcal{H} + \mathcal{K} is the subspace spanned by \mathcal{H} and \mathcal{K} ; writing \mathcal{H} \oplus \mathcal{K} implies that \mathcal{H} and \mathcal{K} are disjoint, and then \mathcal{H} \oplus \mathcal{K} = \mathcal{H} + \mathcal{K} ; and \mathcal{H} \cap \mathcal{K} is the intersection of \mathcal{H} and \mathcal{K} .
- If E_{1}+E_{2}=E is a projection, then (E_{1}+E_{2})^{2}=E^{2}=E=E_{1}+E_{2}, so that the cross-product terms must disappear: E_{1} E_{2}+E_{2} E_{1}=0. \tag{1} If we multiply (1) on both left and right by E_{1} , we obtain \begin{align} & E_{1} E_{2}+E_{1} E_{2} E_{1}=0, \\ & E_{1} E_{2} E_{1}+E_{2} E_{1}=0; \end{align}subtracting, we get E_{1} E_{2}-E_{2} E_{1}=0 . Hence E_{1} and E_{2} are commutative, and (1) implies that their product is zero. (Here is where we need the assumption 1+1 \neq 0 .) Since, conversely, E_{1} E_{2}=E_{2} E_{1}=0 clearly implies (1), we see that the condition is also sufficient to ensure that E be a projection.
Let us suppose, from now on, that E is a projection; by Section: Projections , Theorem 2, \mathcal{M} and \mathcal{N} are, respectively, the sets of all solutions of the equations E z=z and E z=0 . Let us write z=x_{1}+y_{1}=x_{2}+y_{2} , where x_{1}=E_{1} z and x_{2}=E_{2} z are in \mathcal{M}_{1} and \mathcal{M}_{2} , respectively, and y_{1}=(1-E_{1}) z and y_{2}=(1-E_{2}) z are in \mathcal{N}_{1} and \mathcal{N}_{2} , respectively. If z is in \mathcal{M} , E_{1} z+E_{2} z=z , then \begin{align} z&=E_{1}(x_{2}+y_{2})+E_{2}(x_{1}+y_{1})\\ &=E_{1} y_{2}+E_{2} y_{1}. \end{align}Since E_{1}(E_{1} y_{2})=E_{1} y_{2} and E_{2}(E_{2} y_{1})=E_{2} y_{1} , we have exhibited z as a sum of a vector from \mathcal{M}_{1} and a vector from \mathcal{M}_{2} , so that \mathcal{M} \subset \mathcal{M}_{1}+\mathcal{M}_{2} . Conversely, if z is a sum of a vector from \mathcal{M}_{1} and a vector from \mathcal{M}_{2} , then (E_{1}+E_{2}) z=z , so that z is in \mathcal{M} , and consequently \mathcal{M}=\mathcal{M}_{1}+\mathcal{M}_{2} . Finally, if z belongs to both \mathcal{M}_1 and \mathcal{M}_{2} , so that E_{1} z=E_{2} z=z , then z=E_{1} z=E_{1}(E_{2} z)=0, so that \mathcal{M}_{1} and \mathcal{M}_{2} are disjoint; we have proved that \mathcal{M}=\mathcal{M}_{1} \oplus \mathcal{M}_{2} .
It remains to find \mathcal{N} , that is, to find all solutions of E_1z + E_2z = 0 . If z is in \mathcal{N}_{1} \cap \mathcal{N}_{2} , this equation is clearly satisfied; conversely E_{1} z+E_{2} z=0 implies (upon multiplication on the left by E_{1} and E_{2} respectively) that E_{1} z+E_{1} E_{2} z=0 and E_{2} E_{1} z+E_{2} z=0 . Since E_{1} E_{2} z=E_{2} E_{1} z=0 for all z , we obtain finally E_{1} z=E_{2} z=0 , so that z belongs to both \mathcal{N}_{1} and \mathcal{N}_{2} .
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, E_{1}-E_{2} is a projection if and only if 1-(E_{1}-E_{2})=(1-E_{1})+E_{2} is a projection. According to (i) this happens (since, of course, 1-E_{1} is the projection on \mathcal{N}_{1} along \mathcal{M}_{1} ) if and only if (1-E_{1}) E_{2}=E_{2}(1-E_{1})=0, \tag{2} and in this case (1-E_{1})+E_{2} is the projection on \mathcal{N}_{1} \oplus \mathcal{M}_{2} along \mathcal{M}_{1} \cap \mathcal{N}_{2} . Since (2) is equivalent to E_{1} E_{2}=E_{2} E_{1}=E_{2} , the proof of (ii) is complete.
- That E=E_{1} E_{2}=E_{2} E_{1} implies that E is a projection is clear, since E is idempotent. We assume, therefore, that E_{1} and E_{2} commute and we find \mathcal{M} and \mathcal{N} . If E z=z , then E_{1} z=E_{1} E z=E_{1} E_{1} E_{2} z=E_{1} E_{2} z=z, and similarly E_{2} z=z , so that z is contained in both \mathcal{M}_{1} and \mathcal{M}_{2} . The converse is clear; if E_{1} z=z=E_{2} z , then E z=z . Suppose next that E_{1} E_{2} z=0 ; it follows that E_{2} z belongs to \mathcal{N}_{1} , and, from the commutativity of E_{1} and E_{2} , that E_{1} z belongs to \mathcal{N}_{2} . This is more symmetry than we need; since z=E_{2} z+(1-E_{2}) z , and since (1-E_{2}) z is in \mathcal{N}_{2} , we have exhibited z as a sum of a vector from \mathcal{N}_{1} and a vector from \mathcal{N}_{2} . Conversely if z is such a sum, then E_{1} E_{2} z=0 ; this concludes the proof that \mathcal{N}=\mathcal{N}_{1}+\mathcal{N}_{2} .
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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 \mathcal{M} and \mathcal{N} was a direct sum of the given subspaces, in (iii) we stated only that \mathcal{N}=\mathcal{N}_{1}+\mathcal{N}_{2} . Consideration of the possibility E_{1}=E_{2}=E shows that this is unavoidable. Also: the condition of (iii) was asserted to be sufficient only; it is possible to construct projections E_{1} and E_{2} whose product E_{1} E_{2} is a projection, but for which E_{1} E_{2} and E_{2} E_{1} 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.
