Decomposing space of 2x2 matrices over the reals

Mathkid3242
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Homework Statement


Consider the subspace $$W:=\Bigl \{ \begin{bmatrix}
a & b \\
b & a \end{bmatrix} : a,b \in \mathbb{R}\Bigr \}$$ of $$\mathbb{M}^2(\mathbb{R}). $$

I have a few questions about how this can be decomposed.

1) Is there a subspace $$V$$ of $$\mathbb{M}^2(\mathbb{R})$$ such that $$W\oplus V=\mathbb{M}^2(\mathbb{R})$$? If so, what is one?

2) Further, is there a different (i.e., $$\ne V$$) subspace with the same property? And if not, is there a different proper subspace $$U$$ such that $$W+U= \mathbb{M}^2(\mathbb{R})$$?
Which examples, if any, would work for these questions? I haven't made much progress, so seeing explicit examples would help.

Homework Equations

The Attempt at a Solution



I tried considering this. Set $$V= \{A\in \mathbb{M}^2(\mathbb{R}): A^T =-A\}$$, but it didn't work.[/B]
 
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A good start would be to find the dimension of ##W##. Then, you know that the dimension of ##V##, if it exists, should be ##4 - dim(W)##.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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