A special subclass of Hermitian operators is extremely important. This is the positive operators. A positive operator A is defined to be an operator such that for any vector $|v\rangle$, $(|v\rangle, A|v\rangle)$ is a real, non-negative number.
$(|v\rangle, A|v\rangle)$ refers to the inner product/dot product between $|v\rangle$ and $A|v\rangle$
I saw this bit of info in a quantum computing book. It made sense to me at first, Hermitian operators had real eigenvalues, positive operators had real positive eigenvalues. Until it didn't... I just found a positive matrix which isn't Hermitian:
\begin{equation*} A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} \end{equation*}
If we visualize this in "3b1b essence of linear algebra" style, we can see that none of the vectors get flipped around enough, so $(|v\rangle, A|v\rangle)$ is never negative. So it should be a positive operator... But $A^T\neq A$, so it's not Hermitian.
What's wrong here?