Let \(V\) be a finite-dimensional vector space over \(F\), and let \(T\) be a linear operator on \(V\). Suppose that \(T\) commutes with every diagonalizable linear operator on \(V\). Prove that \(T\) is a scalar multiple of the identity operator.
Solution
Every projection \(E\) is diagonalizable: since \(E^2=E\), one has \[ V=\operatorname{range}E\oplus\ker E, \] and \(E\) acts by \(1\) on its range and by \(0\) on its kernel [@HK71]. Therefore the hypothesis implies that \(T\) commutes with every projection.
Fix a nonzero vector \(v\in V\). Extend \(v\) to a basis and let \(E_v\) be the projection onto the line \(Fv\) along the span of the remaining basis vectors. Since \(TE_v=E_vT\), \[ E_v(Tv)=T(E_vv)=Tv, \] so \(Tv\in Fv\). Thus every nonzero vector is an eigenvector of \(T\).
If \(v,w\) are linearly independent, write \[ Tv=av, \qquad Tw=bw, \qquad T(v+w)=c(v+w). \] Linearity gives \[ av+bw=c(v+w), \] and independence implies \(a=b=c\). Comparing every vector with one fixed nonzero vector shows that one scalar \(c\) satisfies \(Tv=cv\) for all \(v\in V\). Hence \[ \boxed{T=cI}. \]