Let $V$ be an inner product space, finitely generated over $\mathbb{C}$, $T\in \operatorname{End}(V)$ that satisfies $TT^*=T^2$, show that $T$ is self-adjoint.
I know that $TT^*$ is positive so has positive square-root, thus the square root is positive definite. But I don't feel like this is a proof. Can anyone prove it without considering square root?
If $Tx=0$, then $T^*x=0$ because $$ \|T^*x\|^2=(T^*x,T^*x)=(TT^*x,x)=(T^2x,x)=0. $$ You are given that $T(T^*-T)=0$, which then forces $T^*(T^*-T)=0$ by the above result, and, hence, also $(T-T^*)(T^*-T)=0$. Therefore, for all $x$, \begin{align} \|(T^*-T)x\|^2&=((T^*-T)x,(T^*-T)x) \\ &=((T-T^*)(T^*-T)x,x) = 0. \end{align} Thus, $\|(T^*-T)x\|=0$ for all $x$. So $T^*x=Tx$ for all $x$, or $T^*=T$.
Choose an orthonormal basis of $V$ such that $T$ is upper-triangular when expressed as a matrix in this basis (we can always do that in a finite-dimensional complex inner product space; by Schur decomposition).
Let us denote the coefficients of that matrix by $T_{ij}$. We need to show that $(T_{ij})_{ij}$ is a diagonal matrix.
Look at the diagonal entries of $TT^*$:
$$(TT^*)_{ii} = \sum_{j=1}^n T_{ij} \overline{T}_{ij} = \sum_{j=1}^n |T_{ij}|^2$$
and the diagonal entries of $T^2$:
$$(T^2)_{ii} = \sum_{j=1}^n T_{ij} T_{ji} = T_{ii}^2$$
because $T$ is upper-triangular. Since $TT^*=T^2$, we have $$\sum_{j=1}^n |T_{ij}|^2 = T_{ii}^2$$
In particular, $T_{ii}^2$ is a positive real number, so $T_{ii}$ is a real number. Thus,
$$\sum_{j=1,j\not=i}^n |T_{ij}|^2 = 0$$
But this implies $T_{ij}=0$ for $j\not=i$, so $T$ is self-adjoint.
