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Cauchy-Schwarz Inequality

  1. Aug 12, 2009 #1
    Hello,

    For two n-dimensional vectors [tex]\mathbf{v}_1\text{ and }\mathbf{v}_2[/tex], what is the Cauchy-Schwarz Inequality:

    1- [tex]|\mathbf{v}_1.\mathbf{v}_2|\leq\|\mathbf{v}_1\|\|\mathbf{v}_2\|[/tex], or

    2- [tex]|\mathbf{v}_1.\mathbf{v}_2|\leq\|\mathbf{v}_1\|+\|\mathbf{v}_2\|[/tex]

    In either case, the equality holds when [tex]\mathbf{v}_1=a\,\mathbf{v}_2[/tex], where a is a positive real constant. Is there any specific way to compute a, or just pick an arbitrary positive real number?

    Regards
     
  2. jcsd
  3. Aug 12, 2009 #2
    I don't believe you can calculate 'a' from the inequality. You would need another route to get it.
     
  4. Aug 12, 2009 #3

    jgens

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    Your second inequality is the "triangle inequality." The Cauchy-Schwarz inequality is [itex]| \langle \mathbf{A},\mathbf{B} \rangle | \leq \|{\mathbf{A}}\| \, \|{\mathbf{B}} \|[/itex]
     
  5. Aug 12, 2009 #4
    the equality holds when v1=av2 for any scalar a.
     
  6. Aug 13, 2009 #5

    uart

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    The first is the CS inequality.


    BTW the second inequality has an error (dot instead or plus). It should read :

    [tex]|\mathbf{v}_1 + \mathbf{v}_2|\leq\|\mathbf{v}_1\|+\|\mathbf{v}_2\|[/tex]

    In which case it's the triangle inequality.
     
  7. Aug 13, 2009 #6
    So, can we write:

    [tex]|\mathbf{w}^H\,\mathbf{h}|^2\leq\|\mathbf{w}\|^2\,\|\mathbf{h}\|^2[/tex]?
     
  8. Aug 14, 2009 #7

    HallsofIvy

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    What they are saying is that equality holds if an only if one vector is a multiple of the other. a could be any real number. It is not a matter of calculating a or a picking a.
     
  9. Aug 14, 2009 #8
    Ok, thank you all guys.
     
  10. Aug 15, 2009 #9
    Could we say, more generally, that a could be any scalar (so that this would hold for any inner product space, over any field)?
     
  11. Aug 15, 2009 #10
    As far as I know, Cauchy proved the inequality for complex vector spaces and Schwarz proved it for polynomial space. In my general linear algebra text, it is proven for real/complex vector spaces. It is a special case of a http://planetmath.org/encyclopedia/CauchySchwartzInequality.html [Broken].
     
    Last edited by a moderator: May 4, 2017
  12. Aug 18, 2009 #11

    HallsofIvy

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    Over any ordered field, yes. That is necessary in order that we be able to say "[itex]\le[/itex]".
     
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