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# Pythagorean theorem in inner product spaces

Pythagorean theorem - 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.

Related:

PythagorasTheorem

Synonym:

Pythagoras theorem in inner product spaces

Type of Math Object:

Theorem

Major Section:

Reference

Groups audience:

## Mathematics Subject Classification

46C05*no label found*

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