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Vectorspace of matrices

  1. Aug 23, 2005 #1
    Hey, I have a quick question. :confused:

    Let [itex]{\cal U}[/itex] be the subspace of [itex]\mathbb{R}_{2x2}[/itex] of all matrices of the form [tex]\left( \begin{array}{ccc}x&-x\\y&z\end{array}\right)[/tex].

    Is it true, that

    [tex]\left( \begin{array}{ccc}x&-x\\y&z\end{array}\right) = x\left( \begin{array}{ccc}1&-1\\0&0\end{array}\right) + y\left( \begin{array}{ccc}0&0\\1&0\end{array}\right) + z\left( \begin{array}{ccc}0&0\\0&1\end{array}\right) = x{\cal M}_1 + y{\cal M}_2 + z{\cal M}_3[/tex]

    So [itex]{\cal B}=\left\{\cal M}_1,{\cal M}_2,{\cal M}_{3} \right\}[/itex] forms a basis for [itex]{\cal U}[/itex]. And that it has dimension 3 after showing that they are independent.


    [tex]{\cal W} = \left( \begin{array}{ccc}a&b\\-a&c\end{array}\right)[/tex]

    \left\{\left( \begin{array}{ccc}1&0\\-1&0\end{array}\right),\left( \begin{array}{ccc}0&1\\0&0\end{array}\right),\left( \begin{array}{ccc}0&0\\0&1\end{array}\right)\right\}

    is a basis for [itex]{\cal W}[/itex] (with three dimensions).

    Any help appreciated. Thanks :smile:
  2. jcsd
  3. Aug 23, 2005 #2


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    Yeah, you just need to show that M1,M2 and M3 are independent.

    The wording of question 2 is wrong, but I take it's an analogous question which has an analogous solution.
  4. Aug 23, 2005 #3
    Yeah sorry, was slack on that bit. What about the intersection? I get some wacky answer. With a basis; the matrices [tex]\left( \begin{array}{ccc}0&0\\0&1\end{array}\right)[/tex] and [tex]\left( \begin{array}{ccc}1&-1\\-1&0\end{array}\right)[/tex]
    Last edited: Aug 23, 2005
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