Pythagorean theorem in inner product spaces

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

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

Proof : As $x\perp y$ one has $\langle x,y\rangle=0$. Then

 $\displaystyle\|x+y\|^{2}$ $\displaystyle=$ $\displaystyle\langle x+y,x+y\rangle$ $\displaystyle=$ $\displaystyle\langle x,x\rangle+\langle x,y\rangle+\langle y,x\rangle+\langle y% ,y\rangle$ $\displaystyle=$ $\displaystyle\|x\|^{2}+\langle x,y\rangle+\overline{\langle x,y\rangle}+\|y\|^% {2}$ $\displaystyle=$ $\displaystyle\|x\|^{2}+\|y\|^{2}\qquad\qquad\qquad\quad\square$

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

Title Pythagorean theorem in inner product spaces PythagoreanTheoremInInnerProductSpaces 2013-03-22 17:32:13 2013-03-22 17:32:13 asteroid (17536) asteroid (17536) 5 asteroid (17536) Theorem msc 46C05 Pythagoras theorem in inner product spaces PythagorasTheorem