glitinosim3

2022-07-14

When using the definition and properties of the inner product, we get the parallelogram law:
$\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}‖x+y{‖}^{2}=\phantom{\rule{mediummathspace}{0ex}}⟨x+y,x+y⟩=⟨x,x⟩+⟨x,y⟩+⟨y,x⟩+⟨y,y⟩\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{0ex}{0ex}}‖x-y{‖}^{2}=\phantom{\rule{mediummathspace}{0ex}}⟨x-y,x-y⟩=⟨x,x⟩-⟨x,y⟩-⟨y,x⟩+⟨y,y⟩$
$‖x+y{‖}^{2}+‖x-y{‖}^{2}\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}2⟨x,x⟩+2⟨y,y⟩\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}2‖x{‖}^{2}+2‖y{‖}^{2}$
as required. But the above does not simplify to Pythagoras' theorem $‖x+y{‖}^{2}=‖x{‖}^{2}+‖y{‖}^{2}$if $x$ and $y$ are orthogonal.

How can we get Pythagoras from the parallelogram law?

Expert

If you are talking strictly about the parallelogram, when the vector are orthogonal (hence $⟨x,y⟩=0$, you have
$||x+y|{|}^{2}=⟨x+y,x+y⟩=⟨x,x⟩+⟨x,y⟩+⟨y,x⟩+⟨y,y⟩=‖x{‖}^{2}+‖y{‖}^{2}$
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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