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Orthogonal complement

  1. Mar 17, 2014 #1
    1. The problem statement, all variables and given/known data
    [itex] G:= [/itex]
    \begin{pmatrix}
    1 & 0 & 0 & 1 \\
    0 & 1 & 1 & 0 \\
    0 & 1 & 1 & 0\\
    1 & 0 & 0 & 1 \\
    \end{pmatrix}

    [itex] B(x,y) = x^{T}Gy[/itex]

    [itex] B: \textit{R}^{4} X \textit{R}^{4} \rightarrow \textit{R}[/itex]

    Find [itex] (\textit{R}^{4})^{\bot} [/itex]

    2. Relevant equations

    [itex] (\textit{R}^{4})^{\bot} = \left\{x | B(x,y)=0,\forall y\in\textit{R}^{4} \right\}[/itex]

    3. The attempt at a solution
    I think I have to solve a linear equation system but I don't know how to set it up.
    I could only find examples where a subspace of the vektor space was given.
    Any hints are much appreciated.
     
  2. jcsd
  3. Mar 17, 2014 #2

    Mark44

    Staff: Mentor

    This is what I would do. Start by writing out what B(x, y) means.

    $$B(x, y) = \begin{bmatrix} x_1 & x_2 & x_3 & x_4\end{bmatrix} \begin{bmatrix}1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4\end{bmatrix} $$

    You want all vectors x in R4 such that B(x, y) = 0 for any y in R4.
     
  4. Mar 17, 2014 #3
    Yes, but wouldn't that give me one equation with 8 unknowns.
    [itex]y_{1}(x_{1}+x_{4})+y_{2}(x_{2}+x_{3})+y_{3}(x_{2}+x_{3})+y_{4}(x_{1}+x_{4}) = 0[/itex]
     
  5. Mar 17, 2014 #4

    Mark44

    Staff: Mentor

    But the expression on the left side has to be identically zero for any choices of the y values, so what conditions does that place on the four x values?
     
  6. Mar 17, 2014 #5
    [itex]x_{1}+x_{4}=0 [/itex] and [itex] x_{2}+x_{3}=0 [/itex]
    Is that what you mean?
     
  7. Mar 17, 2014 #6

    Mark44

    Staff: Mentor

    Yes.
     
  8. Mar 18, 2014 #7
    Then [itex]x_{1} = -x_{4} = -b [/itex], [itex]x_{2} := a [/itex], [itex]x_{3} = -x_{2} = -a [/itex] and [itex]x_{4} := b [/itex]
    [itex] \quad (R^{4})^{\bot}=a\begin{pmatrix}
    0 \\
    1 \\
    -1 \\
    0 \\
    \end{pmatrix}+ b\begin{pmatrix}
    -1 \\
    0 \\
    0 \\
    1 \\
    \end{pmatrix} [/itex]
    Have I got it right?
     
  9. Mar 18, 2014 #8

    Mark44

    Staff: Mentor

    Looks good.
     
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