Let $v$ and $w$ be elements of an inner product space. Prove that $||v + w||^2 = ||v||^2 + ||w||^2$ if and only if $v$, and $w$ are orthogonal.
I'll answer my own question below
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1Since there's no way of getting privileges other than reputation, I've resorted to answering my own questions especially since most of my answers are unaccepted. I'd like an explanation for any downvotes please, and tell me what I should do instead – Cotton Headed Ninnymuggins Nov 16 '22 at 04:35
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Your post (in its original form) is a PSQ = Problem Statement Question. These are generally not acknowledged here, as they do not reflect any effort of your own, or hint at your motivation. Furthermore, the command-style "Prove that ..." is not well received by me, certainly, and many more amongst the Math SE community, for sure. – Hanno Jan 11 '25 at 12:45
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Note: "i.p." is to mean inner-product, and have others have noted, I’m assuming we’re only dealing with $\mathbb{R}$: $$ \begin{aligned} ||v+w||^2 &= \langle v+w,v+w\rangle\\ &= \langle v,v\rangle +\langle v,w\rangle +\langle w,v\rangle +\langle w,w\rangle \\ &= \langle v,v\rangle +2\langle v,w\rangle+\langle w,w\rangle &&(\text{since } \langle v,w\rangle=\langle w,v\rangle \text{ when i.p. exists}) \end{aligned} $$
And
$$ ||v||^2+||w||^2=\langle v,v\rangle +\langle w,w\rangle $$
Now
$$ ||v + w||^2 = ||v||^2 + ||w||^2\\ \implies\langle v,v\rangle +\color{red}{2\langle v,w\rangle } +\langle w,w\rangle=\langle v,v\rangle +\langle w,w\rangle $$
This is only true iff $2\langle v,w \rangle=0\implies \langle v,w \rangle=0$ which is the definition of $2$ vectors being orthogonal.
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3If $\left< u,w \right> =ib$ with $b\ne 0$,then $\left< u,w \right> +\left< w,u \right> =ib-ib=0$. But $u$ and $w$ are not orthogonal. – fusheng Nov 16 '22 at 04:42
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@fusheng Is my proof simply incomplete or is the original statement false? – Cotton Headed Ninnymuggins Nov 16 '22 at 06:20
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@fusheng and I thought that if $\langle u,w \rangle$ was a defined inner product, that $\langle u,w \rangle = \langle w,u \rangle$? – Cotton Headed Ninnymuggins Nov 16 '22 at 06:22
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Over $\mathbb{C}$, it's instead $\langle u,w \rangle = \overline{\langle w,u \rangle}$. – Bruno B Nov 16 '22 at 07:55
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Your proof is true for an inner product space over $\mathbb{R}$. Bur if it is over $\mathbb{C}$, it is not true. – fusheng Nov 16 '22 at 09:33
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@fusheng ohhh ok. I’m assuming we’re dealing with $\mathbb{R}$ because that’s all we’ve done so far in Linear Algebra, and the question is to prove the statement true. – Cotton Headed Ninnymuggins Nov 16 '22 at 16:34