On my linear algebra exam, I was required to find the vectors which span $\mathrm{ker(T)}$, where the linear transformation $\mathrm{T}$ was given. The official answers are $\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}$ or $\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}$.
I wrote my answer as $ \begin{bmatrix} 0 & a\\ -a & 0 \end{bmatrix},a\in \mathbb{R} $
I did not get credit for this problem, as the official solutions says we had to write the vectors which span $\mathrm{ker(T)}$ and not the set $\mathrm{ker(T)}$.
I realise my answer is not completely right, as when $\mathrm{a} = 0$ the vector no longer spans $\mathrm{ker(T)}$. But other than that, is there something else I'm missing? If I had written $\mathrm{a} \in \mathbb{R} - \{0\}$, would my answer then be correct? Also, what is the difference between $\mathrm{ker(T)}$ and the "set" $\mathrm{ker(T)}$ ?
By definition, the $\ker{T}$ is a set. It is defined as follows: $$\ker{T} = \{x\in E: Tx = 0\}$$ where $T:E\to V.$
Your answer would have been correct if you were asked to just state the kernel. You would write something like this: $$\ker{T} = \left\{\begin{bmatrix} 0 & a\\ -a & 0 \end{bmatrix},a\in \mathbb{K}\right\}$$ However notice that $$\ker{T} = \left\{\begin{bmatrix} 0 & a\\ -a & 0 \end{bmatrix},a\in \mathbb{K}\right\} = \operatorname{Span}\left(\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}\right).$$ Here the $\operatorname{Span}(v) = \{\lambda v:\lambda \in \mathbb{K}\}$, where $\mathbb{K}$ is the field over which you have defined your vector space. So when they ask you for the vectors that span the kernel you have to esentially tell them the vectors whose scalar mutiples form the kernel. You could also find a basis for the kernel and that would be just fine too.