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?
Best Answer
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 $\,W_1\,$