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LINEAR ALGEBRA: Show that |a x| = |a| |x| for all real numbers a

  1. Oct 31, 2006 #1
    How does one prove this statement?

    I have no idea how to start. Can someone help?

    Maybe it has something to do with that Cauchy-Schwartz inequality?
     
  2. jcsd
  3. Oct 31, 2006 #2
    I guess using bilinearity of dot product spaces will work, right?

    Since (a x) * y = a (x * y) = x * (ay)
     
  4. Oct 31, 2006 #3

    quasar987

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    I don't think bilinearity of dot product space is something someone struggling to prove |a x| = |a| |x| knows about! :bugeye:

    You could use the definition of |.|, namely that

    [tex]|ax|=\left\{ \begin{array} {c} -ax \ \ \mbox{for} \ \ ax<0 & +ax \ \ \mbox{for} \ \ ax\geq 0 \end{array}[/tex]

    and treat the three cases a<0, a>0, a=0 separetely.
     
  5. Oct 31, 2006 #4
    [tex]|a|=\left\{ \begin{array} {c} -a \ \ \mbox{for} \ \ a<0 & +a \ \ \mbox{for} \ \ a\geq 0 \end{array}[/tex]

    [tex]|x|=\left\{ \begin{array} {c} -x \ \ \mbox{for} \ \ x<0 & +x \ \ \mbox{for} \ \ x\geq 0 \end{array}[/tex]

    Does this somehow "prove" that |a x| = |a| |x| [itex]\in[/itex] R?

    If not, what would count as "proof"?
     
    Last edited: Oct 31, 2006
  6. Oct 31, 2006 #5

    quasar987

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    I don't understand your question. My post was meant as a hint to the OP.
     
  7. Oct 31, 2006 #6
    I think he means, is what you posted sufficient to constitute a proof?
     
  8. Oct 31, 2006 #7

    Office_Shredder

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    EDIT: Wrong absolute value. I should read the thread
     
    Last edited: Oct 31, 2006
  9. Oct 31, 2006 #8

    HallsofIvy

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    If you know about "bilinearity of dot product" then surely you understood that the question was about the lengths of vectors, not about absolute value? "|ax|= -ax for ax< 0" makes no sense because the left side of the equation is a number and the right side is a vector. Also, there is no order defined on a vector space.
     
  10. Oct 31, 2006 #9

    Galileo

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    Isn't |ax|=|a||x| by definition of a norm on a vector space?
     
  11. Oct 31, 2006 #10

    radou

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    Let a be a real number, and [tex]\vec{v} \in V^2(O)[/tex], for simplicity.

    [tex]\left|a\right|\cdot\left|\vec{v}\right|=\left|a\right|\sqrt{v_{x}^2+v_{y}^2}=\sqrt{\left|a\right|^2(v_{x}^2+v_{y}^2)}=\sqrt{a^2(v_{x}^2+v_{y}^2)}=\sqrt{(av_{x})^2+(av_{y})^2}=\left|a \cdot \vec{v}\right|[/tex]. Could this be considered as a proof?
     
  12. Oct 31, 2006 #11

    quasar987

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    I didn't realise that post #2 was by the OP himself, lol. And for some reason I was certain this thread was about real numbers, despite its name LINEAR ALGEBRA. Sorry for all the confusion VinnyCee!
     
  13. Nov 1, 2006 #12

    HallsofIvy

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    That is part of the general definition. I suspect that this problem is based on
    [tex]|ai+ bj+ ck|= \sqrt{a^2+ b^2+ c^2}[/tex]
    for R3 or
    [tex]|ai+ bj|= \sqrt{a^2+ b^2}[/tex]
    for R2. In that case radou's post is exactly what he needs.
     
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