Proving $\langle HX, Y \rangle = \langle X, HY \rangle$ for Hermitian matrices

Question

Let $X, Y \in \mathbb{C}^n$. Let $\langle \cdot, \cdot \rangle$ be the standard Hermitian innerproduct, defined as $\langle X, Y \rangle = \overline{X}^T Y$. I want to prove the following assertion:

Proposition: A complex ($n \times n)$-matrix $H$ is Hermitian if and only if \begin{align*} \langle HX, Y \rangle = \langle X, HY \rangle. \end{align*}

I proved one direction as follows. Suppose $H$ is Hermitian. Then we have \begin{align*} \langle HX, Y \rangle = (\overline{HX})^T Y = \overline{X}^T \overline{H}^T Y = \overline{X}^T H Y = \langle X, HY \rangle. \end{align*} But I'm not sure how to prove the converse. How to show on the basis of the above property that $H$ is Hermitian?

Edit: can I just write out the LHS and the RHS, this would give me $\overline{X}^T \overline{H}^T Y$ and $\overline{X}^T H Y$, and then conclude that $H = \overline{H}^T$, i.e. $H$ is Hermitian?

Answer 1

Hint: Subtracting your LHS and RHS expressions, we have $$ X^*(H - H^*)Y = 0 $$ for every pair of vectors $X,Y \in \Bbb C^n$. Use this to deduce that $H - H^* = 0$.

Answer 2

Hint. The equation has to be true for all $X$ and $Y$, thus in particular when $X$ and $Y$ are the standard basis vectors $\mathbf e_i$ and $\mathbf e_j$.