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Skew symmetric matrix question

  1. Sep 7, 2008 #1
    Let A in n x n real matrix.
    For every x in R^n we have <Ax,x>=0 where < , > is scalar product.
    prove that A^t=-A (A is skew symmetric matrix)
     
  2. jcsd
  3. Sep 7, 2008 #2

    tiny-tim

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    Hi TTob! :smile:

    Hint: just write out <Ax,x>=0 in terms of Aij etc …

    then jiggle it around a bit! :smile:

    What do you get?
     
  4. Sep 7, 2008 #3
    Thank you.

    note [tex]x=(x_1,...,x_n)[/tex] and [tex]A=(a_{ij})[/tex].
    then [tex](Ax)_i=[/tex] [tex]\sum_{\substack{0\leq j\leq n}} a_{ij}x_j[/tex]
    then
    [tex]<Ax,x>=[/tex] [tex]\sum_{\substack{0\leq j\leq n \\ 0\leq i\leq n}} a_{ij}x_i x_j=0[/tex]

    when you put [tex]x=e_i[/tex] you get [tex]a_{ii}=0[/tex] for all i.
    when you put [tex]x=e_i+e_j[/tex] you get [tex]a_{ii}+a_{ij}+a_{ji}+a_{jj}=0[/tex] for all i,j .

    so [tex]a_{ji}=-a_{ij}[/tex] for all i,j and then [tex]A^t=-A[/tex].
     
  5. Sep 7, 2008 #4

    tiny-tim

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    Hi TTob! :smile:

    Yes … or if you want to avoid using components …

    [tex]0\ =\ <A(x+y),x+y>\ -\ <Ax,x>\ -\ <Ay,y>\ \ =\ \ <Ax,y>\ +\ <Ay,x>\ \ =\ \ <(A + A^T)x,y>[/tex] :smile:
     
  6. Sep 7, 2008 #5

    HallsofIvy

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    Or use the fact that <Ax, y>= <x, ATy>. (That is a more general definition of "adjoint")

    From that, <Ax, x>= <x, ATx>. If A is skew-symmetric, <x, AT x>= -<Ax, x> so <Ax, x>= -<Ax, x> and <Ax, x>= 0.
     
  7. Sep 7, 2008 #6

    tiny-tim

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    Hi HallsofIvy! :smile:

    erm …
    I think you've proved the transpose of the original theorem! :wink:
     
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