Direct Sums, Projections, and Idemsquare Operators

I have often seen a projection operator on a vector space* defined as one that is equal to its square (under composition). That's...just wrong. A projection is an operator that...well...projects, preferably orthogonally but that's not necessary. Here we define projections correctly and state as a theorem that they are idemsquare. (*I think all this works more generally for abelian groups.) Conversely, we prove that any linear operator \(T\) satisfying \(T^2=T\) is a projection operator.

Let us start by recalling how projections arise and how they are defined. The context is the direct-sum decomposition of a vector space. (All this is easily adapted for abelian groups or even R-modules for a commutative ring R. Even so, projections most commonly arise in the vector space context.)

As usual, a vector space \(W\) is the direct sum of two subspaces \(U\) and \(V\) iff every vector \(w\) in \(W\) is uniquely the sum of a vector \(u\) in \(U\) and a vector \(v\) in \(V\). We write \(W=U\oplus V\) to indicate that \(W\) is the direct sum of \(U\) and \(V\). It is easy to show that, for any subspaces \(U\) and \(V\) of \(W\),\(W=U\oplus V\) iff \(W=U+V\) and \(U\cap V=\left\{\vec0\right\}\). Because of the uniqueness in the relation among \(w\), \(u\), and \(v\) in the definition of direct sum, \(u\) and \(v\) are actually functions of \(w\) (for fixed \(U\) and \(V\)). We call \(u\) the projection of \(w\) onto \(U\) along \(V\) and \(v\) the projection of \(w\) onto \(V\) along \(U\). We write \(u=\mbox{ proj}_{U,V}w\) and \(v=\mbox{ proj}_{V,U}w\). It is easy to show that

these projection functions are linear operators on \(W\),
the range of the operator \(\mbox{proj}_{U,V}\) is \(U\)
its kernel is \(V\).
...and similarly for \(\mbox{ proj}_{V,U}\).

We may now state the...

Theorem. (Fundamental ~ of Projections on Vector Spaces) For all vector spaces \(W\) and any linear operator \(T\) on \(W\), equivalent are

(1) \(W=U\oplus V\) and \(T=\mbox{ proj}_{U,V}\) where \(U=\mbox{ ran }T\) and \(V=\ker T\) (i.e., \(T\) is the projection onto its range along its kernel).
(2) \(T\) is a projection operator onto some subspace of \(W\) along some (other) subspace (i.e., \(T\) is some projection operator).
(3) \(T^2=T\), i.e., T is idemsquare.

Proof. Suppose \(W\) is a vector space, \(T\) is a linear operator on \(W\), and \(U\) and \(V\) are subspaces of \(W\).

\((1\Rightarrow2)\) Suppose \(W=U\oplus V\) and \(T=\mbox{ proj}_{U,V}\) where \(U=\mbox{ ran }T\) and \(V=\ker T\). Then \(T\) is the projection onto its range along its kernel, whence Then \(T\) is manifestly the projection onto some subspace of \(W\) along some other, namely, its range and kernel.

\((2\Rightarrow3)\) Suppose \(T\) is the projection operator onto some subspace of \(W\) along some other subspace. By definition of “projection”, there exist such subspaces \(U\) and\(V\) of \(W\) that \(W=U\oplus V\) and \(T=\mbox{ proj}_{U,V}\). It is easy to check that \(V=\ker T\). For any \(w\in W\), write \(w=u+v\) with \(u\in U\) and \(v\in V\). Then \(u=Tu=T(u+v)=Tw\). Thus, \(T^2w=T((Tw))= Tu=Tw\), whence \(T^2=T\).

\((3\Rightarrow1)\) Suppose \(T^2=T\). Let \(U=\mbox{ ran }T\) and \(V=\ker T\). We show first that \(W=U\oplus V\), i.e., that \(W=U+V\) and that every vector \(w\in W\) is uniquely the sum of a vector \(u\in U\) and a vector \(v\in V\).

- \((U+V=W)\) consider \(w\in W\) . Let \(u = Tw\) and \(v = w – Tw\). Then \(w = u + v\). Vector \(u\) is in \(T\)'s range (being the image of \(w\)) and \(v\) is in the kernel because \(Tv=T(w-Tw)=Tw-T^2(w)=Tw-Tw=\vec0\), where we used the hypothesis \(T^2=T\). Thus, \(U+V=W\).
- \((U\oplus V=W)\) Now suppose we can write \(w\) as a sum from \(U\) and \(V\) in two ways: \(w=u+v\) and \(w=u'+v'\). \(T\) is the identity on its range (because \(T^2=T\)) and so \(Tu = u\) and \(Tu’ = u’\). Then
  - \(Tw=T(u+v)=Tu+Tv=Tu+\vec0=Tu=u\) and
  - \(Tw=T(u'+v')=Tu'+Tv'=Tu'+\vec0=Tu'=u'\).

whence \(u = u’\), whence \(v = v’\), whence the sum is unique. It is now legal to define the projection \(\mbox{proj}_{U,V}\). Recall that \(Tw = u\) and \(w=u+v\) by hypothesis, so we get \(u=\mbox{ proj}_{U,V}\). Thus, \(T=\mbox{ proj}_{U,V}\), whence \(U=\mbox{ ran }T\) and \(V=\ker T\).

We have established that the three statements are equivalent, as promised.

Thus, being a projection is equivalent to being idemsquare, although that should not be the definition of a projection.