Especially important for our purposes is another connection between direct sums and linear transformations.
Definition 1. If \(\mathcal{V}\) is the direct sum of \(\mathcal{M}\) and \(\mathcal{N}\) , so that every \(z\) in \(\mathcal{V}\) may be written, uniquely, in the form \(z=x+y\) with \(x\) in \(\mathcal{M}\) and \(y\) \(\mathcal{N}\) , the projection on \(\mathcal{M}\) along \(\mathcal{N}\) is the transformation \(E\) defined by \(Ez = x\) .
If direct sums are important, then projections are also, since, as we shall see, they are a very powerful algebraic tool in studying the geometric concept of direct sum. The reader will easily satisfy himself about the reason for the word "projection" by drawing a pair of axes (linear manifolds) in the plane (their direct sum). To make the picture look general enough, do not draw perpendicular axes!
We skipped over one point whose proof is easy enough to skip over, but whose existence should be recognized; it must be shown that \(E\) is a linear transformation. We leave this verification to the reader, and go on to look for special properties of projections.
Theorem 1. A linear transformation \(E\) is a projection on some subspace if and only if it is idempotent, that is, \(E^{2}=E\) .
Proof. If \(E\) is the projection on \(\mathcal{M}\) along \(\mathcal{N}\) , and if \(z=x+y\) , with \(x\) in \(\mathcal{M}\) and \(y\) in \(\mathcal{N}\) , then the decomposition of \(x\) is \(x+0\) , so that \[E^{2} z=E E z=E x=x=E z.\] Conversely, suppose that \(E^{2}=E\) . Let \(\mathcal{N}\) be the set of all vectors \(z\) in \(\mathcal{V}\) for which \(E z=0\) ; let \(\mathcal{M}\) be the set of all vectors \(z\) for which \(E z=z\) . It is clear that both \(\mathcal{M}\) and \(\mathcal{N}\) are subspaces; we shall prove that \(\mathcal{V} = \mathcal{M} \oplus \mathcal{N}\) . In view of the theorem of Section: Direct sums , we need to prove that \(\mathcal{M}\) and \(\mathcal{N}\) are disjoint and that together they span \(\mathcal{V}\) .
If \(z\) is in \(\mathcal{M}\) , then \(E z=z\) ; if \(z\) is in \(\mathcal{N}\) , then \(E z=0\) ; hence if \(z\) is in both \(\mathcal{M}\) and \(\mathcal{N}\) , then \(z=0\) . For an arbitrary \(z\) we have \[z=E z+(1-E) z.\] If we write \(E z=x\) and \((1-E) z=y\) , then \[E x=E^{2} z=E z=x,\] and \[E y=E(1-E) z=E z-E^{2} z=0,\] so that \(x\) is in \(\mathcal{M}\) and \(y\) is in \(\mathcal{N}\) . This proves that \(\mathcal{V} = \mathcal{M} \oplus \mathcal{N}\) , and that the projection on \(\mathcal{M}\) along \(\mathcal{N}\) is precisely \(E\) . ◻
As an immediate consequence of the above proof we obtain also the following result.
Theorem 2. If \(E\) is the projection on \(\mathcal{M}\) along \(\mathcal{N}\) , then \(\mathcal{M}\) and \(\mathcal{N}\) are, respectively, the sets of all solutions of the equations \(E z=z\) and \(E z=0\) .
By means of these two theorems we can remove the apparent asymmetry, in the definition of projections, between the roles played by \(\mathcal{M}\) and \(\mathcal{N}\) . If to every \(z=x+y\) we make correspond not \(x\) but \(y\) , we also get an idempotent linear transformation. This transformation (namely, \(1-E\) ) is the projection on \(\mathcal{N}\) along \(\mathcal{M}\) . We sum up the facts as follows.
Theorem 3. A linear transformation \(E\) is a projection if and only if \(1-E\) is a projection; if \(E\) is the projection on \(\mathcal{M}\) along \(\mathcal{N}\) , then \(1-E\) is the projection on \(\mathcal{N}\) along \(\mathcal{M}\) .