Finding ranks and nullities of linear maps I am confused about ranks, nullities and bases of the kernel.
From what I understand the rank is the dimension of a vector space generated by a matrix.
How would I do the following examples?

Find the ranks and nullities of the following linear maps $T \colon U \to V$, and find bases of the kernel and image of $T$ in each case.
(i) $U = \mathbb{R}^4$, $V = \mathbb{R}^4$, $T(\alpha, \beta, \gamma, \delta) = (\alpha - \gamma, \gamma-\delta, \alpha-\beta, \beta-\delta)$;
(ii) $U = \mathbb{R}[x]_{\leq 5}$, $V = \mathbb{R}[x]_{\leq 5}$ (polynomials of degree at most $5$ over $\mathbb{R}$), $T(f) = f'''$ (third derivative of $f \in U$).

(Source)
 A: Note that for the second exercise a basis for $V=\Bbb R[x]_{\le 5}$ is given by: $(1,x,x^2,x^3,x^4,x^5)$. To calculate $\ker$ and $ImT$ consider the matrix taking this basis. $$A=\begin{pmatrix}0&0&0&6&0&0\\0&0&0&0&24&0\\0&0&0&0&0&60\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\end{pmatrix}$$ Now you can calculate $\ker A$ and $Im A$
A: For (i), let's start with the kernel. Suppose $T(a,b,c,d)=(0,0,0,0)$. Then $(a-c,c-d,a-b,b-d)=(0,0,0,0)$ which implies $a=b=c=d$. Therefore the kernel is the
subspace generated by $\{(a,a,a,a)\}$ for some $a\in\mathbb{R}$. This has dimension (nullity) equal to $1$. 
Now by the rank-nullity theorem, the rank is $3$. If you haven't seen this theorem then the matrix of $T$ is equal to
$$\begin{pmatrix}1&0&-1&0\\0&0&1&-1\\1&-1&0&0\\0&1&0&-1\end{pmatrix}$$
(try to calculate this yourself or at least convince yourself of why it's true)
You can row reduce this to get the rank, which should equal $3$ like we've said.
For (ii), you just need to compute the matrix of $T$ again.
