Let \(T\) be a linear operator on \(V\) which commutes with every projection operator on \(V\). What can you say about \(T\)?
Solution
One must have \[ \boxed{T=cI} \] for some scalar \(c\).
1. Every one-dimensional subspace is \(T\)-invariant.
Let \(0\ne v\in V\) and put \(L=\operatorname{span}(v)\). Extend \(v\) to a basis of \(V\) and let \(M\) be the span of the remaining basis vectors. Then \[ V=L\oplus M, \] so there is a projection \(E\) onto \(L\) along \(M\) [@HK71]. Since \(TE=ET\), for any \(x\in L\) one has \(Ex=x\) and therefore \[ E(Tx)=T(Ex)=Tx. \] Thus \(Tx\in\operatorname{range}E=L\). In particular there is a scalar \(c_v\) such that \[ Tv=c_vv. \] Hence every nonzero vector is an eigenvector of \(T\).
2. All these eigenvalues are equal.
If \(\dim V\le1\), every linear operator already has the form \(cI\). Assume \(\dim V\ge2\) and take linearly independent vectors \(v,w\). By step 1, \[ Tv=av, \qquad Tw=bw, \qquad T(v+w)=c(v+w) \] for some scalars \(a,b,c\). Linearity also gives \[ T(v+w)=av+bw. \] Therefore \[ (a-c)v+(b-c)w=0. \] Independence of \(v,w\) implies \(a=b=c\).
Now fix one nonzero vector \(v_0\). For any nonzero \(u\), either \(u\) is a scalar multiple of \(v_0\), in which case its eigenvalue is the same by linearity, or \(u\) is independent of \(v_0\), in which case the preceding argument gives the same value. Thus a single scalar \(c\) satisfies \(Tv=cv\) for every \(v\in V\). Hence \(T=cI\).