Is the Intersection of Two Matrix Subspaces Nontrivial?

[gazzo]

Hey, I have a quick question.

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

Is it true, that

\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

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

Also,

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

So
<br /> \left\{\left( \begin{array}{ccc}1&amp;0\\-1&amp;0\end{array}\right),\left( \begin{array}{ccc}0&amp;1\\0&amp;0\end{array}\right),\left( \begin{array}{ccc}0&amp;0\\0&amp;1\end{array}\right)\right\}<br />

is a basis for {\cal W} (with three dimensions).

Any help appreciated. Thanks

 

[Galileo]

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.

 

[gazzo]

Yeah sorry, was slack on that bit. What about the intersection? I get some wacky answer. With a basis; the matrices \left( \begin{array}{ccc}0&amp;0\\0&amp;1\end{array}\right) and \left( \begin{array}{ccc}1&amp;-1\\-1&amp;0\end{array}\right)