Prove that $Trace(XX^*) = 0$ We have that $H$ is an invariant subspace of a normal operator $L$ and that $P$ is the orthogonal projection to $H$. We want to prove that for $X = PL(I-P)$, we have  $Trace(XX^*) = 0$.
I got that $X = PL - LP$ so $X^* = L^*P^* - P^*L^*$ so then I got the product. I then tried to use the property that trace is invariant under cyclic permutations and that $LL^* = L^*L$, but I couldn't get anywhere. Am I going in the right direction here? What are other properties that might be useful?
 A: Suppose we are working in a finitely dimensional inner product space $V$ [since we are going to prove some trace being $0$]. 


*

*$L|_H$ is diagonalizable. 


Restrict $L, P$ to $H$, then $P|_H$ is actually the identity on $H$. Since $L$ is normal, $L$ is diagonalizable, and then $L|_H$ is diagonalizable as well. Thus there is  an eigenbasis $(v_j)_1^r$ of $H$ w.r.t. to $L|_H$, and $Lv_j = c_j v_j$ for $j \leqslant r$. 


*$L^*|_{H ^\perp}$ is diagonalizable. 


Now by the definition of adjoints, $H^{\perp}$ is invariant under $L^*$. Since $L^*$ is also normal, $L^*$ is diagonalizable. Thus $L^*|_{H^\perp}$ is diagonalizable as well, so there is an eigenbasis $(v_j)_{r+1}^n$ w.r.t. $L^*|_{H^\perp}$. 


*$L|_{H^\perp}$ is also diagonalizable, thus $L$ is a diagonalizable with $L|_H$ and $L|_{H^\perp}$ being diagonal under some basis of $V$. 


Suppose $L^*v_j = c_j v_j$. Then for $j \geqslant r+1$
$$
\newcommand\Norm[1]{\left\Vert #1 \right\Vert}
\Norm {Lv_j - \overline {c_j}v_j}^2 = (Lv_j|Lv_j) + |c_j|^2 \Norm{v_j}^2 - (Lv_j|\overline {c_j} v_j) - (\overline {c_j}v_j|Lv_j) = (v_j|LL^* v_j) + |c_j|^2 \Norm{v_j}^2 - c_j(v_j|L^*v_j) - \overline {c_j}(L^*v_j|v_j) = \Norm {L^*v_j}^2 + |c_j|^2 \Norm {v_j}^2 - c_j\overline {c_j}\Norm {v_j}^2 - \overline {c_j} c_j \Norm{v_j}^2 = 0, 
$$
hence $Lv_j = \overline {c_j}v_j$ for $j \geqslant r+1$. Therefore $(v_j)_1^n$ is an eigenbasis of $V$ w.r.t. $L$ [since $V = H \oplus H^{\perp}$], and $Pv_j = v_j$ when $j \leqslant r$, $Pv_j = 0$ when $j \geqslant r+1$. 


*$X = LP-PL =O$. 


Thus 
$$
Xv_j =(LP-PL)v_j=
\begin{cases}
Lv_j - P(c_j v_j) = c_jv_j - c_j Pv_j, & j \leqslant r,\\
L 0 - P(\overline{c_j} v_j) = 0-\overline {c_j}Pv_j, & j\geqslant r+1
\end{cases} = 0, 
$$
hence $X = O$, then $\mathrm {Tr}(XX^*)=0$.
Note: if $\mathrm {Tr}(XX^*)=0$, then $X = O$, since as a matrix $X= [x_{j,k}]$, the trace is just $\sum_{j,k} |x_{j,k}|^2 =0$, i.e. $x_{j,k}=0$ for all $j,k$. Thus we aim at proving $X=O$. 
