I have a homework question that I am very unsure if my solution is sufficient, mainly in the converse of the proof, since I feel like I need to use $\mu_T(x)$ somewhere in my proof.
Question: Assume $T$ is an operator on the finite-dimensional inner product space $V$ and the minimum polynomial of $T$ is $\mu_T(x)=x^2-x$. Let $U=E_1$ be the subspace of fixed vectors, i.e. $T(u)=u, \forall u\in U$. Let $W=\ker(T)$. Prove that $T$ is self-adjoint if and only if $W=U^\perp$.
My attempt: Assume $T$ is self-adjoint. Thus, we have that $T=T^*$. Let $x\in U,y\in W$. We will show that $\langle x,y\rangle=0$. Since $x$ is fixed by $T$, we have
$$\langle x,y\rangle=\langle T(x),y\rangle=\langle x,T(y)\rangle=\langle x,0\rangle=0$$
Thus since $x,y$ were arbitrary, we have shown that every vector in $U$ is orthogonal to every vector in $W$ and thus $W=U^\perp$.
Conversely, assume $W=U^\perp$. Again, let $x\in U,y\in W$. We can see
$$0=\langle x,y\rangle=\langle T(x),y\rangle=\langle x,T^*(y)\rangle$$
additionally we have
$$0=\langle x,0\rangle=\langle x,T(y)\rangle=\langle T^*(x),y\rangle$$
Thus we have that $\langle T(x),y\rangle=\langle x,T(y)\rangle$ for all $x\in U,y\in W$. Since $U$ is the eigenspace related to $\lambda=1$, $U$ is a subspace of $V$. Thus, we know that $U+W=U+U^\perp=V$, and thus this solution works for all vectors in $V$. So $T=T^*$.
I feel like my solution to the converse is very naive and that I am assuming too much so I feel I am missing a big part of this question. If it is incorrect, some advice on how to start it would be appreciated.
Best Answer
You have to take arbitrary elements $x,y \in V$ and show that $\langle Tx,y\rangle =\langle x,Ty\rangle$. For this write $x=u+w,y=u'+w'$ with $u,u' \in U$ and $v,v' \in W$. We get $\langle Tx,y\rangle =\langle u,u'+w'\rangle$ (because $T(u+w)=u+0=u$) and this becomes $\langle u,u'\rangle$ (because $\langle u,w'\rangle =0)$. Similarly, $\langle x,Ty\rangle=\langle u+w,u'\rangle=\langle u,u'\rangle$. This finishes the proof.