Is the positive operator $TT^*-T^*T$ on a finite complex inner product space is a zero operator? Suppose the vector space is a complex finite-dimensional inner product space $V$, and the operator $TT^*-T^*T$ is a positive operator on it. Can this information be sufficient to conclude that $TT^*-T^*T$ is a zero operator on $V$?
I think maybe it can guarantee that $TT^*-T^*T=0 $ (or $T$ is normal),but I can't prove this

Here are just some variants of Stephen's answer in the matrix form (just for references by other readers):
In terms of matrices, we know that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ for every two $n\times n$ complex matrices. In particular, we always have $\operatorname{tr}(A^*A)=\operatorname{tr}(AA^*)$, which gives $\operatorname{tr}(A^*A-AA^*)=0$.
If $A^*A-AA^*$ is positive semi-definite, then its eigenvalues are non-negative whence $\operatorname{tr}(A^*A-AA^*)\ge0$. Now the inequality is attained, which means that all the eigenvalues of $A^*A-AA^*$ should be zero!
Of course we cannot conclude that a matrix is zero if all of its eigenvalues are zero, but note that $A^*A-AA^*$ is Hermitian whence a normal matrix. It means that $A^*A-AA^*$ is unitarily similar to the zero matrix, hence $A^*A-AA^*=\mathbf{0}$ and
$A^*A=AA^*$, i.e., $A$ is normal.
