Equality of inner product of two complex vectors with equal norms

Question

Suppose $x,y \in \mathbb{C}^m$ (vectors in $m$ dimensional complex space).

How can I prove that $x^Hy = y^Hx$ if $\|x\|_2 = \|y\|_2$ where $(\cdot)^H$ is the Hermitian form (conjugate transpose) and $\|\cdot\|_2$ is the vector 2-norm? Thx!

Answer 1

Let $x = i, y=1$. Then $x^H = -i, y^H = 1$. $x^H y = -i \cdot 1 = -i$. $y^H x = 1 \cdot i = i$. $\|x\|_2 = 1 = \|y\|_2 $. $-i \neq i$. So this contradicts your proposition.

Try to prove instead that $x^H y = \overline{y^H x}$ where $\overline{c}$ is the complex conjugate of $c$. You do not need any assumptions on the norms of $x,y$.