glitinosim3

Answered

2022-07-14

When using the definition and properties of the inner product, we get the parallelogram law:

$\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Vert x+y{\Vert}^{2}=\phantom{\rule{mediummathspace}{0ex}}\u27e8x+y,x+y\u27e9=\u27e8x,x\u27e9+\u27e8x,y\u27e9+\u27e8y,x\u27e9+\u27e8y,y\u27e9\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{0ex}{0ex}}\Vert x-y{\Vert}^{2}=\phantom{\rule{mediummathspace}{0ex}}\u27e8x-y,x-y\u27e9=\u27e8x,x\u27e9-\u27e8x,y\u27e9-\u27e8y,x\u27e9+\u27e8y,y\u27e9$

Adding these two expressions:

$\Vert x+y{\Vert}^{2}+\Vert x-y{\Vert}^{2}\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}2\u27e8x,x\u27e9+2\u27e8y,y\u27e9\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}2\Vert x{\Vert}^{2}+2\Vert y{\Vert}^{2}$

as required. But the above does not simplify to Pythagoras' theorem $\Vert x+y{\Vert}^{2}=\Vert x{\Vert}^{2}+\Vert y{\Vert}^{2}$if $x$ and $y$ are orthogonal.

How can we get Pythagoras from the parallelogram law?

$\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Vert x+y{\Vert}^{2}=\phantom{\rule{mediummathspace}{0ex}}\u27e8x+y,x+y\u27e9=\u27e8x,x\u27e9+\u27e8x,y\u27e9+\u27e8y,x\u27e9+\u27e8y,y\u27e9\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{0ex}{0ex}}\Vert x-y{\Vert}^{2}=\phantom{\rule{mediummathspace}{0ex}}\u27e8x-y,x-y\u27e9=\u27e8x,x\u27e9-\u27e8x,y\u27e9-\u27e8y,x\u27e9+\u27e8y,y\u27e9$

Adding these two expressions:

$\Vert x+y{\Vert}^{2}+\Vert x-y{\Vert}^{2}\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}2\u27e8x,x\u27e9+2\u27e8y,y\u27e9\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}2\Vert x{\Vert}^{2}+2\Vert y{\Vert}^{2}$

as required. But the above does not simplify to Pythagoras' theorem $\Vert x+y{\Vert}^{2}=\Vert x{\Vert}^{2}+\Vert y{\Vert}^{2}$if $x$ and $y$ are orthogonal.

How can we get Pythagoras from the parallelogram law?

Answer & Explanation

persstemc1

Expert

2022-07-15Added 18 answers

If you are talking strictly about the parallelogram, when the vector are orthogonal (hence $\u27e8x,y\u27e9=0$, you have

$||x+y|{|}^{2}=\u27e8x+y,x+y\u27e9=\u27e8x,x\u27e9+\u27e8x,y\u27e9+\u27e8y,x\u27e9+\u27e8y,y\u27e9=\Vert x{\Vert}^{2}+\Vert y{\Vert}^{2}$

You can also see it "geometrically", by noticing that this means that when $x$ and $y$ are orthogonal, you have a rectangle, with both diagonals of same length, that is

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

$||x+y|{|}^{2}=\u27e8x+y,x+y\u27e9=\u27e8x,x\u27e9+\u27e8x,y\u27e9+\u27e8y,x\u27e9+\u27e8y,y\u27e9=\Vert x{\Vert}^{2}+\Vert y{\Vert}^{2}$

You can also see it "geometrically", by noticing that this means that when $x$ and $y$ are orthogonal, you have a rectangle, with both diagonals of same length, that is

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

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