= Solution
For a <linear map> $T:V\to W$, its rank is $\dim\operatorname{im}T$ and its nullity is $\dim\ker T$. The <rank-nullity theorem> states that, when $V$ is finite-dimensional,
$$
\dim V=\dim\ker T+\dim\operatorname{im}T.
$$
To prove it, take a <basis> $v_1,\ldots,v_k$ of $\ker T$ and extend it to a <basis>
$$
v_1,\ldots,v_k,v_{k+1},\ldots,v_n
$$
of $V$. The <vectors> $Tv_{k+1},\ldots,Tv_n$ span $\operatorname{im}T$. They are also linearly independent: if $\sum_{j=k+1}^na_jTv_j=0$, then $\sum_{j=k+1}^na_jv_j\in\ker T$, and independence of the chosen <basis> forces every $a_j$ to vanish. Thus the rank is $n-k$ and the nullity is $k$.
For the given subspace, row reduction of the coefficient <matrix> gives
$$
\begin{pmatrix}
1&0&3&0&43/3\\
0&1&3&0&-5/3\\
0&0&0&1&-11/3
\end{pmatrix}.
$$
Taking $x_3=s$ and $x_5=t$ therefore gives
$$
(x_1,x_2,x_3,x_4,x_5)
=s(-3,-3,1,0,0)+\frac t3(-43,5,0,11,3).
$$
Hence
$$
\boxed{\dim W=2},
\qquad
\boxed{\{(-3,-3,1,0,0),(-43,5,0,11,3)\}}
$$
is a <basis> of $W$.
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