If $M$ is the vector space of $2\times 2$ real matrices, then I can show that

$$ \{A \in M \mid A^\mathrm{T}=-A \} $$

is a subspace of $M$, since 

$$ \left[ \begin{array}{cc} x & z \\ -z & y \end{array} \right]+\left[ \begin{array}{cc} x' & z' \\ -z' & y' \end{array} \right] = \left[ \begin{array}{cc} x+x' & z+z' \\ -(z+z') & y+y' \end{array} \right] $$

and for some $\lambda\in\mathbb{R}$ 

$$ \lambda\left[ \begin{array}{cc} x & z \\ -z & y \end{array} \right] = \left[ \begin{array}{cc} \lambda x & \lambda z \\ -\lambda z & \lambda y \end{array} \right] $$

But I'm not sure if I'm correct on finding the dimension and a basis of the subspace:

$$ \left[ \begin{array}{cc} x & z \\ -z & y \end{array} \right] = x\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]+y\left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right]+z\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] $$

This makes me think that a basis is made up of 

$$ \left(\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right],\left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right],\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] \right) $$

and so the dimension is three. Is that right?