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numbertheory
 Post subject: qx dot qy
PostPosted: Wed, 23 May 2012 10:48:43 UTC 
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prove
(Qx) · (Qy) = x · y

given
(Qx+Qy) · (Qx+Qy) = ||x+y||
||Qx||=||x||
||Qy||=||y||

ie can't do this - <Qx,Qy>=(Qx)^(T)Qy=(x^T)(Q^T)Qy
or use the 1/4th identity
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 Post subject: Re: qx dot qy
PostPosted: Wed, 23 May 2012 18:59:45 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12097
Location: Austin, TX
numbertheory wrote:
prove
(Qx) · (Qy) = x · y

given
(Qx+Qy) · (Qx+Qy) = ||x+y||
||Qx||=||x||
||Qy||=||y||

ie can't do this - <Qx,Qy>=(Qx)^(T)Qy=(x^T)(Q^T)Qy
or use the 1/4th identity


Whoa, slow down. What is Q?

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