I understand that if you have a linear transformation from $U$ to $V$ with, say, $\operatorname{dim} U = 3$, $\operatorname{rank} T = 2$, then the set of points that map onto the $0$ vector will lie along a straight line, and therefore $\operatorname{nullity}T = 1$.
Can anyone offer an intuitive explanation of why this is always true?
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I like saulspatz' answer because it is very hands-on. I would like to offer another perspective, one based on the fact that linear transformations are characterized by their ranks, up to choice of bases in domain and codomain. The key is in this proposition:
Proposition. Suppose $T\colon V\to W$ is a linear transformation with $\dim V = n$ and $\dim W = m$ and $\operatorname{rank} T = r \leqslant m$. Then there are bases $v_1,\dots,v_n$ for $V$ and $w_1,\dots,w_m$ for $W$ such that the matrix for $T$ with respect to these bases is $$ \mathcal M(T,v_1,\dots,v_n,w_1,\dots,w_m) = \begin{pmatrix} I_{r\times r} & 0_{r\times n-r} \\ 0_{m-r\times r} & 0_{m-r\times n-r}\end{pmatrix}, $$ where $I_{r\times r}$ is the $r\times r$ identity matrix, and the various $0_{\ast\times\ast}$ are the zero matrices of the corresponding dimensions. As a quick corollary of the proposition, we can read off of the matrix for $T$ in these bases that \begin{align*} \operatorname{rank}T &\stackrel{\text{def}}{=} \dim \operatorname{image}T = r, \\ \operatorname{nullity}T &\stackrel{\text{def}}{=} \dim \ker T = n-r, \end{align*} and hence gain the rank-nullity theorem: $$ \dim V = n = r + (n-r) = \operatorname{rank}T + \operatorname{nullity}T. $$ For a quick proof of the proposition, keeping things as "coarse" as possible for intuition's sake, because the rank of $T$ is $r$, take vectors $v_1,\dots,v_r$ in $V$ such that $w_1 = T(v_1),\dots,w_r = T(v_r)$ span the image of $T$. Extend $v_1,\dots,v_r$ to a basis $v_1,\dots,v_r,v_{r+1},\dots,v_n$ for $V$ and $w_1,\dots,w_r$ to a basis $w_1,\dots,w_r,w_{r+1},\dots,w_m$ for $W$. With respect to these bases, we quickly determine $$ \mathcal M(T,v_1,\dots,v_n,w_1,\dots,w_m) = \begin{pmatrix} I_{r\times r} & \ast \\ 0_{m-r\times r} & \ast\end{pmatrix}. $$ Because the rank of $T$ is $r$, and the first $r$ columns of the matrix for $T$ are linearly independent, we determine that (possibly after some row and column operations) the two $\ast$'s in the above matrix for $T$ have to be the zero matrices of appropriate dimensions, hence the proposition.
I like this question. Let me take a shot at it. I think it's best to think of the rank as the dimension of the range (or image).
Consider first a nonsingular transformation $T$ on an $n-$dimensional vector space. We know that the rank is $n$ and the nullity $0$, so the theorem holds in this case. $T$ maps a basis to a basis. Suppose we modify $T$ by mapping the first vector in the basis to $0$. Call the new transformation $T_1$. Clearly, the nullity of $T$ is $1$. What is the rank? In the image of $T$ one of the basis vectors collapses to $0$ when we go to the image of $T_1,$ so the image of $T_1$ has dimension $n-1$ and the theorem hold in this case.
Now continue the process. If $T_2$ is the same as $T_1$ except that the second basis vector is mapped to $0$, then the nullity will be $2$, and the image will be of dimension $n-2$, because again, one dimension collapses.
Of course, we can continue until we arrive at $T_n=0$ and the theorem always holds.
I hope this makes intuitive sense to you.