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Let $V = M^{2\times 2}(\bf F),$

$W_1 =\left\{\begin{bmatrix}a & b \\c & a\end{bmatrix}\in V\;:\; a, b, c\in F\right\}$

and

$W_2 =\left\{ \begin{bmatrix}0 & a \\-a & b\end{bmatrix}\in V\;:\; a, b, \in F\right\}$

Prove that $W_1$ and $W_2$ are subspaces of $V$ and find the dimensions of $\,W_1\,,\, W_2\,,\, W_1+W_2\,,\, W_1\cap W_2\,$.

My attempt: Clearly, $W_1$ is of dimension $3$ since it has three independent components, and $W_2$ is of dimension $2$ since it only has $2$. However, does this mean $W_1+W_2$ will have $\dim = 3$ since there will be three independents in total? How do I prove that?

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    Yo$u$ meant a set of matrices? Yes, you can say that, yet that is not what you wrote in your question's heading.2012-09-28

1 Answers 1

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Since $\,\dim (W_1+W_2)=\dim W_1+\dim W_2-\dim(W_1\cap W_2)\,$ , it is probably wiser to try to find first the dimension of the intersection:

$A=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}\in W_1\cap W_2\Longleftrightarrow 0=\alpha=\delta\,\,,\,\,\beta=-\gamma$

and from the above clearly the dimension of the intersection is $\,1\,$ . Now complete the answer.

Added: $\,W_1\,$ is closed under multiplication by scalar:

$k\begin{pmatrix}a&b\\c&a\end{pmatrix}:=\begin{pmatrix} ka&kb\\kc&ka\end{pmatrix}$

and we can easily see the right hand matrix belongs to $\,W_1\,$ as its main diagonals' elements are equal, as required from any element in $\,W_1\,$

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    I'll add some hints to this in my original answer. Wait for it.2012-09-28