Let $T$ be the linear transformation represented by the matrix
$ \left( \begin{array}{ccc} 1 & 1 & 0 & 3 \\ 1 & 1 & 1 & 5 \\ 2 & 2 & 1 & 8 \end{array} \right) $
One can easily calculate that the image of this as a map from $\mathbb{R}^4\to\mathbb{R}^3$ is 3.
Call the above matrix $A$. Now consider the space $V$ of linear maps $B$ from $\mathbb{R}^2\to\mathbb{R}^4$ satisfying $AB=0$. Plainly $B$ is in $V$ iff the image of $B$ is in the kernel of $T$.
What is the dimension of $V$? Directly calculating, I get 4. However, it seems that one could argue that the dimension is the dimension of the kernel of $T$, which is $2$. What is the flaw in reasoning?