Relating $N(A)$ and dimension

Question

Let $B$ be a $2 * 3$ matrix. Suppose that we know that $N(B) = span \left\{ \begin{pmatrix} 1 \\ 2 \\ 3\\ \end{pmatrix}\right\}$.

Would it be correct to assume by the rank-nullity theorem, that since the rank of B is at most 2, then with the information above, the rank of B would have to be 1? Can we then extend this to hold for $3*4, 4*5, 5*6...$ matrices?

Answer 1

For your first question, see JDZ’s answer. I’ll point out, though, that you seem to be confusing the two vector spaces involved. You’re right that the rank of $B$ is a most $\dim\mathbb R^2$, but the total dimension in the Rank-Nullity theorem is that of the map’s domain, i.e., of $\mathbb R^3$.

To answer your second question, this is exactly what the Rank-Nullity Theorem tells us. An $n\times m$ matrix $A$ represents a linear map $T:\mathbb R^m\to\mathbb R^n$ via left-multiplication, so $\operatorname{rank}(A)+\operatorname{nullity}(A)=\dim\operatorname{im}T+\dim\ker T = \dim\mathbb R^m=m.$

Answer 2

$B$ determines a linear map $\mathbb{R}^{3}\rightarrow\mathbb{R}^{2}$ via left multiplication. If $\dim\,\textrm{Ker}(B)=1$ as you suppose, then $3=\dim\,\mathbb{R}^{3}=\dim\,\textrm{Ker}(B)+\dim\,\textrm{Im}(B)=1+\dim\,\textrm{Im}(B)$ implies $\dim\,\textrm{Im}(B)=2$ by the rank-nullity theorem.