Pythagorean theorem in inner product spaces

- Let $X$ be an inner product space (over $ℝ$ or $ℂ$) and $x,y\in X$ two orthogonal vectors. Then

$${\parallel x+y\parallel }^2={\parallel x\parallel }^2+{\parallel y\parallel }^2.$$

Proof : As $x⟂y$ one has $⟨x,y⟩=0$. Then

${\parallel x+y\parallel }^2$$=$$⟨x+y,x+y⟩$

$=$$⟨x,x⟩+⟨x,y⟩+⟨y,x⟩+⟨y,y⟩$

$=$${\parallel x\parallel }^2+⟨x,y⟩+\overline{⟨x,y⟩}+{\parallel y\parallel }^2$

$=$${\parallel x\parallel }^2+{\parallel y\parallel }^2\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{\square }$

$Remark-$ This theorem is valid (with the same proof) for spaces with a semi-definite inner product.