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Sign considering in question

  1. Oct 28, 2011 #1
    1)
    [tex]q(x_{1,}x_{2,}x_{3})=[/tex]
    [tex]x_{1}^{2}+5x_{2}^{2}+26x_{3}^{2}+2x_{1}x_{2}+10x_{1}x_{3}+6x_{2}x_{3}[/tex]

    [tex]V=\{x=(x1,x2,x3)\in R^{3}:q(x)=0\}[/tex]

    check if V is a subspace of [tex]R^{3}[/tex] and

    find the basis of V?
    how i tried:
    i diagonolized it the representative by rows and columns and i see
    that
    q is semi positive(having all posotive numbers except one zero)
    [tex]A=\left(\begin{array}{ccc}1 & 1 & 5\\1 & 5 & 3\\5 & 3 & 26\end{array}\right)=>A=\left(\begin{array}{ccc}1 & 0 & 0\\0 & 4 & 0\\0 & 0 & 0\end{array}\right)[/tex]
    my prof said that because q(x) is semi positive then V is subspace
    but why??
    and he didnt even looked that q(x)=0 it could be q(x)>=0
    he made his desition without considering if q(x)=0 or q(x)>=0.
    and when i asked him about the basis of V
    i looked at it when q(x)=0 as kernel of A
    but he said it worng and didnt say why
    ?
    so how to find the basis
    ?
     
    Last edited: Oct 28, 2011
  2. jcsd
  3. Oct 29, 2011 #2
    I would say if the diagonal A' and A are related by the change of basis matrix T such that
    T^-1 A' T= A, then you need to apply this transformation to the vector (0, 0, 1), so your basis would be {T^-1 (0, 0, 1) T}
    That's why you diagonalized your matrix in the first place. Suppose you are acting your function on (x',y',z'), then your result is x'^2+4y'^2, so anything that has only z will be mapped to zero.
     
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