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Cross product in arbitrary field

  1. May 12, 2015 #1
    Let [itex]\mathbb{F}[/itex] be an arbitrary field, and let [itex]a,b\in\mathbb{F}^3[/itex] be vectors of the three dimensional vector space. How do you prove that if [itex]a\times b=0[/itex], then [itex]a[/itex] and [itex]b[/itex] are linearly dependent?

    Consider the following attempt at a counter example: In [itex]\mathbb{R}^3[/itex]

    [tex]
    \left(\begin{array}{c}
    1 \\ 4 \\ 2 \\
    \end{array}\right)
    \times\left(\begin{array}{c}
    2 \\ 3 \\ 4 \\
    \end{array}\right)
    = \left(\begin{array}{c}
    10 \\ 0 \\ -5 \\
    \end{array}\right)
    [/tex]

    holds. Since [itex]5[/itex] is a prime number, [itex]\mathbb{Z}_5[/itex] is a field. In [itex](\mathbb{Z}_5)^3[/itex]

    [tex]
    \left(\begin{array}{c}
    [ 1] \\ [ 4] \\ [ 2] \\
    \end{array}\right)
    \times \left(\begin{array}{c}
    [2] \\ [ 3] \\ [ 4] \\
    \end{array}\right)
    = \left(\begin{array}{c}
    [ 0] \\ [ 0] \\ [ 0] \\
    \end{array}\right)
    [/tex]

    This might look like a counter example to the claim. One might consider the possibility that perhaps the claim is true for example when [itex]\mathbb{F}[/itex] is a subfield of [itex]\mathbb{C}[/itex], but not in general?

    A closer look reveals that the counter example attempt does not work, because

    [tex]
    [ 2] \left(\begin{array}{c}
    [ 1] \\ [ 4] \\ [ 2] \\
    \end{array}\right)
    = \left(\begin{array}{c}
    [ 2] \\ [ 3] \\ [ 4] \\
    \end{array}\right)
    [/tex]

    holds in [itex](\mathbb{Z}_5)^3[/itex]. Finding a counter example is difficult, and it seems that the claim is true after all. I only know how to prove the claim using determinants when [itex]\mathbb{F}[/itex] is a subfield of [itex]\mathbb{C}[/itex].
     
  2. jcsd
  3. May 12, 2015 #2

    micromass

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    Assume that ##a## and ##b## are linearly independent in the field ##\mathbb{F}^3##, then you can expand them to a basis ##(c,a,b)##. Because of linear algebra, this means that the matrix formed by ##a##, ##b## and ##c## is invertible and thus has nonzero determinant. But this determinant is ##c\cdot (a\times b)##. Since this is nonzero, it implies that ##a\times b## must be nonzero.
     
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