I saw the following proof for row rank = column rank using rank nullity:
From rank nullity theorem: $\dim(\ker(f)) + \dim(\text{Im}(f)) = \dim(V)$.
$\dim(\text{Im}(f)) = \dim(V) - \dim(\ker(f))$.
By definition, the dimension of the image of $f$ is the column rank. Similarly, by definition, the row rank is given by the difference between the dimension of the space $V$ and the dimension of the kernel of $f$.
Thus, column rank = row rank. I get why $\dim(\text{Im}(f)) =$ column rank but not why $\dim(V) - \dim(\ker(f)) =$ row rank.
Why (if at all) is the definition of row rank as the number of linearly independent rows in the matrix equivalent to dim(V) - dim(ker(f))?
It's because the row space (span of the row vectors) is orthogonal to the nullspace.
To see why, let $A\in \mathbb{R}^{m\times n}$ be a matrix and let $\text{Im}(A^T)$ denote the row space. If $x \in \ker(A)$, we have that $$Ax = 0 \implies a_1\cdot x = \cdots = a_m\cdot x = 0$$ where $a_i$ is a row vector of $A$. This implies that $\text{Im}(A^T) = \ker(A)^\perp$. Since $\ker(A)$ is a subspace, we have $$\dim(\ker(A)) + \dim(\ker(A)^\perp) = \dim(\ker(A)) + \dim(\text{Im}(A^T)) = n $$ and so $$\dim(\text{Im}(A^T)) = n - \dim(\ker(A)).$$