Proving matrices W1 and W2 are subspaces and finding their dimensions

Question

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

$$W1 =\left\{\begin{bmatrix}a & b \\c & a\end{bmatrix}\in V\;:\; a, b, c\in F\right\}$$

and

$$W2 =\left\{ \begin{bmatrix}0 & a \\-a & b\end{bmatrix}\in V\;:\; a, b, \in F\right\}$$

Prove that W1 and W2 are subspaces of $V$ and find the dimensions of $\,W1\,,\, W2\,,\, W1+W2\,,\, W1\cap W2\,$.

My attempt: Clearly, W1 is of dimension 3 since it has three independent components, and W2 is of dimension 2 since it only has 2. However, does this mean W1+W2 will have dim = 3 since there will be three independents in total? How do I prove that?

Answer 1

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 $\,W_1\,$