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
On
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.
Let $e_i$, $1 \le i \le n$, be an orthonormal basis for $V$. Note that $\sum_{i=1}^n {\|T e_i\|}^2_2$ is the sum of the squares of the absolute values of the entries of the matrix representing $T$. Thus $$ \sum_{i=1}^n {\|T e_i\|}^2_2 = \sum_{i=1}^n {\|T^* e_i\|}^2_2 .$$ Now, if $TT^* - T^*T$ is a positive operator, this means that ${\|T^*x\|}_2^2 - {\|Tx\|}_2^2 \ge 0$ for all $x$. Thus ${\|T^*e_i\|}_2 = {\|Te_i\|}_2$ for all $1 \le i \le n$. Since $e_1$ is an arbitrary norm one vector, from this we can see that ${\|T^* x\|}_2 = {\|T x\|}_2$ for every $x \in V$.
Thus $S = TT^* - T^*T$ satisfies $\langle x, S x \rangle = 0$, and if $S$ is a positive operator, this implies $S = 0$.
This is essentially the same proof as using the trace, but unwrapped so that trace is never explicitly mentioned.