Proving an operator is Self-adjoint using the Spectral Theorem

Question

If we let $f: V \rightarrow V$ be a normal linear operator on a complex inner product space $V$ such that $f^4 = f^3$, can we use the spectral theorem to prove $f$ is self-adjoint and $f^2 = f$?

I worked with the matrix representation of $f$, made the change of basis to diagonal form, $D$, and noticed that $D^4 - D^3 = 0$ implies that $D$ is either the identity or the zero transformation - I think this means that the eigenvalues are real, but cannot figure out how to prove that this is a self-adjoint linear operator. Any hints given would be appreciated.

Answer 1

I'm assuming $V$ is finite-dimensional because you mention matrices.

Because $f : V \rightarrow V$ is normal, then $f$ has a matrix representation with respect to an orthonormal basis, $f = U^{\star}DU$, where $U^{\star}U=I=UU^{\star}$ and $D$ is a diagonal matrix. So the diagonal entries $d_{jj}$ must satisfy $d_{jj}^{3}(d_{jj}-1)=0$, which means that the diagonal entries are either $0$ or $1$ and, hence, $D^{\star}=D=D^{2}$. So $f$ inherits these properties: $$ f^{\star} = (U^{\star}DU)^{\star}=U^{\star}D^{\star}U=U^{\star}DU=f,\\ f^{2} = U^{\star}DUU^{\star}DU=U^{\star}D^{2}U=U^{\star}DU=f. $$ Therefore $f$ is an orthogonal projection, i.e., $f^{2}=f=f^{\star}$.