This is my approach:
For a solution to exist, $b$ has to be in column space. $b$ is an $m \times 1$ vector (as necessitated by the rules of matrix multiplication). This means it is a vector in $R^m$, so, for $b$ to be in column space, column space must also be $m$-dimensional. i.e. $r = m$, and $m – r$ (dimension of left null space) $= 0$?
However, I feel like I might be missing something in my understanding, and any clarity on this would be much appreciated.
Best Answer
You need to say "for every $b\in\Bbb R^m$" in your question, but, yes. Just as the nullspace of $A$ is the orthogonal complement of the row space of $A$, the left nullspace of $A$ is the orthogonal complement of the column space. For every $b$ to be in the column space is to say that the column space is all of $\Bbb R^m$.