trace identities for the functional calculus

Question

I'm sorry if this is a trivial question, but I cannot convince myself of why $\text{Tr}\,f(EFE)=\text{Tr}\,f(FEF)$ for projections $E$ and $F$ on a Hilbert space and a (say, continuous or Borel) function $f.$ Thank your for your help!

Answer 1

Expanding upon the comment by @JimmyK4542 .

Let $X=EFE$ and $Y=FEF$, then $$ \eqalign { \rm{tr}(X^k) &= \rm{tr}((EF)^kE) \cr &= \rm{tr}(E(EF)^k) \cr &= \rm{tr}((EF)^k) \cr &= \rm{tr}((FE)^k) \cr &= \rm{tr}(Y^k) \cr } $$ Now expand the function as a Taylor series and take its trace $$ \eqalign { \rm{tr}(\rm{f}(X)) &= \rm{tr}\bigg(\sum^n_{k=0} c_k X^k\bigg) \cr &= \sum^n_{k=0} c_k \rm{tr}(X^k) \cr &= \sum^n_{k=0} c_k \rm{tr}(Y^k) \cr &= \rm{tr}(\rm{f}(Y)) \cr } $$