I posted a question, which I have attempted all except for c. Could someone tell me if I did the question correctly? And how do I begin (c). Thanks, there is a PDF file attached.
in part (a) you have a couple of mistakes: in proving that W is closed under addition that $\displaystyle -a+b$ should be changed to $\displaystyle -(a+b).$ also $\displaystyle \sigma$ is not in $\displaystyle W.$ it's in your base field.
finally in order to show that a non-empty set is a subspace you do not need to prove that it contains $\displaystyle 0.$ unfortunately some instructors give this wrong idea to some students.
for part (b), the basis you found is wrong! you should at least see that none of the elements of the set that you think is a basis basically belongs to $\displaystyle W.$ anyway, a basis of $\displaystyle W$
has only one element. for example $\displaystyle \left \{ \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix} \right \}.$
for part (c) first see that an $\displaystyle n \times n,$ matrix $\displaystyle A=[a_{ij}],$ with real (or complex) entries, is antisymmetric iff $\displaystyle a_{ii}=0$ for all $\displaystyle 1 \leq i \leq n$ and $\displaystyle a_{ij}=-a_{ji}$ for all $\displaystyle 1 \leq i \neq j \leq n.$ can you
see the general form of $\displaystyle A$? if not, try to do it for n = 3 first. now it should be easy to show that a basis for $\displaystyle n \times n$ antisymmetric matrices has $\displaystyle \frac{n(n-1)}{2}$ elements. what are they?
The reason why we often do prove that the 0 vector is in the set is simply to prove it is non-empty- and 0 is typically easiest to work with. Here, you are asked to show that the set of anti-symmetric 2 by 2 matrices is a subspace and I would disagree with millerst- you are NOT given that it is non-empty, you need to show that. Of course, you could do that as well by showing that $\displaystyle \begin{bmatrix}0 & 1 // -1 & 0\end{bmatrix}$ is in the set as by showing that $\displaystyle \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$ is in the set.