Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sign considering in question

  1. Oct 28, 2011 #1

    [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
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook