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How to prove $ |\langle u,v\rangle | \leq ||u||||v||$

Note: I have given this many attempts so don't downvote due to lack of effort, refer to edit history for evidence of said effort

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    $\begingroup$ The last second "identity" is wrong $\endgroup$ – Shuchang May 29 '14 at 6:01
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    $\begingroup$ For an extreme example, use $u=(1,0)$ and $v=(0,1)$. $\endgroup$ – André Nicolas May 29 '14 at 6:02
  • $\begingroup$ Why the 2 downvotes? I am trying to understand... $\endgroup$ – user142198 May 29 '14 at 6:03
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    $\begingroup$ @user142198: your bigger mistake is not understanding the sigma operator and how square roots interact with sums. $(\sum x_i)(\sum y_i)\neq\sum x_iy_i$ and $\sqrt{\sum x_i}\neq\sum\sqrt{x_i}$. $\endgroup$ – symplectomorphic May 29 '14 at 6:11
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    $\begingroup$ I knew what you were "getting at". However, in math it's a good idea to say what you mean and mean what you say because math is hard, and us readers can easily get confused even if you say everything right, so there's no need to add to our confusion by saying things wrong. $\endgroup$ – bof May 29 '14 at 6:57
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Your $$\sqrt{\sum_{i=1}^n u_i u_i} \sqrt{\sum_{i=1}^n v_i v_i} = \sqrt{\sum_{i=1}^n u_i^2 v_i^2}$$ and $$ \sqrt{\sum_{i=1}^n u_i^2 v_i^2} = \sum_{i=1}^n u_i v_i$$ are both wrong.

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  • $\begingroup$ On reflection they are both silly. Is that the only scenario that is wrong? $\endgroup$ – user142198 May 29 '14 at 6:05

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