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 \(\mathcal{K}\) .

1. 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.

1. 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.
2. 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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