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Vectorspace of matrices

  • Thread starter gazzo
  • Start date
  • #1
175
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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.

Also,

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

So
[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\}
[/tex]

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

Any help appreciated. Thanks :smile:
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,989
6
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.
 
  • #3
175
0
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:

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