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

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    Orthogonality
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Vectors X and Y in R^n are orthogonal if and only if the equality ||X+Y|| = ||X-Y|| holds true. The proof involves using the definition of the norm, where the squared norms of the sums and differences of the vectors are expressed in terms of their dot products. If X and Y are orthogonal, their dot product is zero, leading to the equality of the squared norms. Conversely, if the norms are equal, it results in the dot product being zero, confirming the orthogonality of the vectors. This establishes the necessary and sufficient condition for orthogonality in linear algebra.
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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.
 
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