This may sound pedantic but maybe it helps me clarify some misunderstanding. The setup is as follows: Let $A$ be a complex $n\times n$ matrix which is self-adjoint, i.e. $A^*=A$, wherein $*$ denotes the conjugate of the transpose of $A$. Now to state that $A$ is in addition positive semi-definite one has to check that \begin{equation}\label{pos}\langle x, Ax \rangle \geq 0\end{equation} holds.
Now, I read some time series book and in the text the author checks for positive semi-definitness of a matrix $H$ via (notation adapted from the book): $$aHa^*\geq 0.$$ I interpret the latter as $$\langle H^Ta, a\rangle \geq 0$$ which is equivalent to: $$\langle a, \overline{(\left((H^T)^T\right)}a\rangle = \langle a,\overline{H}a \rangle \geq 0$$ However this is testing if the conjugate of $H$ is positive semidefinite. Where am I wrong?
Some of the confusion is that $\langle x, Ax\rangle$ assumes column vectors whereas $aHa^*$ assumes row vectors, so even $H^Ta$ is nonsensical (unless we define $H^T(a)$ to be the matrix product $aH^T$...? it gets messy quickly). But the main problem is that your $H^T$ should have been $H^*$.
Let me first assume that everything is column vectors to make it algebraically simpler. Then it would say $a^*Ha$. This is equal to $\langle Ha, a\rangle$ by definition, which is also equal to $\langle a, H^* a\rangle$ and $\langle a,Ha\rangle$ (the last one is only true because $H=H^*$). Nowhere will $H^T$ this appear, which was your mistake.
With row vectors we should probably write either $aHa^* = \langle a, aH\rangle$ or $aHa^* = \langle aH, a\rangle$, but I don't know if either convention is generally accepted. In any case, they are equivalent in this case because $\langle aH, a\rangle = \langle a, aH\rangle$ since $H$ is self-adjoint.