Let V be an n-dimensional inner product space over C and T ∈ L(V ). Suppose that $T^2 = T^∗T$. Prove that T is self-adjoint.
My idea is since $T^*T$ is positive, then all the eigenvalues of it should be nonnegative. Also, since $T^2=T^*T$, then this is also true for $T^2$. Then suppose $T$ has a eigenvalue $\lambda$, then $T^2$ should have eigenvalue $\lambda^2\geq0$. This forces all the eigenvalues for $T$ must be real, since it's impossible to have $(a+bi)^2\geq0$ with $b\neq 0$. Thus, now I think can prove some other things to prove $T$ is self-adjoint.
If $T$ is normal, then $T$ is self-adjoint
By Schur theorem, there exists an orthonormal basis such that $T$ is upper triangular. If I can prove all the entries above the diagonal of $M(T)$ are zero, then $T$ is self-adjoint
However, unfortunately, I can prove none of them. Also, I was able to solve a similar problem
If $TT^*=T^2$ then $T$ is self-adjoint
This question enable me to say that if $Tv=0$ then $T^*v=0$ by using norm $$ \|T^*v\|^2=\langle T^*v,T^*v\rangle=\langle TT^*v,v\rangle=\langle T^2v,v\rangle=0$$
And by $\dim$ null $T$ = $\dim$ null $T^*$, I can prove null $T$ = null $T^*$ which gives $T$ is normal, but this seems not useful in this problem, or probably I missed something
Anyway, any help is appreciated. Thank!
If you were able to prove
$$\tag1TT^*=T^2\ \implies\ T=T^*$$
and you have $T^*T=T^2$, then taking adjoints you have $T^*T=(T^*)^2$. Letting $S=T^*$, this is $$ SS^*=S^2.$$ Now you know by $(1)$ that $S=S^*$, so $T=T^*$.
Using more explicitly that you are in a finite-dimensional setting, here is another argument. You have $$ (T-T^*)^2=T^2+T^{*2}-T^*T-TT^*=T^*T-TT^*. $$ Multiplying by $i$ and taking the trace, $$ \operatorname{Tr}\big[[i(T-T^*)]^2\big]=0 $$ So the operator $i(T-T^*)$ is selfadjoint and its square is zero; so it is zero. Then $T-T^*=0$.