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C.'s question at Yahoo! Answers (orthogonality).

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello C.

Using the definition of norm:

$ \left\|x+y\right\|^2=(x+y)\cdot (x+y)=x\cdot x+x\cdot y+y\cdot x+y\cdot y= \left\|x\right\|^2+ \left\|y\right\|^2+2\;x\cdot y\\
\left\|x-y\right\|^2=(x-y)\cdot (x-y)=x\cdot x-x\cdot y-y\cdot x+y\cdot y= \left\|x\right\|^2+ \left\|y\right\|^2-2\;x\cdot y\\
$
If $x$ and $y$ are orthogonal, then $x\cdot y=0$ as a consequence $\left\|x+y\right\|^2=\left\|x-y\right\|^2$ or equivalently $\left\|x+y\right\|=\left\|x-y\right\|$.

On the other hand if $\left\|x+y\right\|=\left\|x-y\right\|$, then $\left\|x+y\right\|^2=\left\|x-y\right\|^2$ which implies $4\;x\cdot y=0$ or equivalently $x\cdot y=0$ that is, $x$ and $y$ are orthogonal.