According to the textbook regarding Jordan canonical form, I was encountered by the theorem below.
When we define $A^{g}$ for some square martix $A$ such that $A^{g}= G^{-1}A^{*}G$, where $G$ is some positive-definitive hermitian matrix, if $A^{g}A = AA^{g} $holds, A is called G-normal matrix. And under the circumstance, the equation below holds.
$(Ax, Ax) = (x, A^{g}Ax) $
I don't understand why the equation holds. I would be happy if you could provide the proof.
I'll assume that the inner product $( \cdot, \cdot)$ is defined using $G$ as
$$ (x,y) = x^{*} \cdot G \cdot y. $$ Then
$$ (x, A^{g}Ax) = x^{*} G A^{g} Ax = x^{*} A^{*} G Ax = (Ax)^{*} G (Ax) = (Ax,Ax).$$