Is there a more general version of $row rank(A) = column rank(A)$? $A$ is a $m\times n$ matrix in field $F$ There's a conclusion in matrix that, given $A$ a $m\times n$ matrix in field $F$, one has $$row\, rank(A) = column\,rank(A)$$
Since linear algebra conclusions are sometimes related to more general ones in abstract algebra, I'm wondering, is there a more general version of this conclusion?

Edit: let me explain what is "more general".
For example in linear matrix we have: $A * B = I_n \Rightarrow B * A = I_n$.
In abstract algebra we have that in a group, if there is a "left identity", and every element has a "left inverse", then the left identity is also right identity, and left inverse is also right inverse.
The conclusion in abstract algebra is more general and powerful as it can be applied to matrix product as a group, also can be applied to other cases.

Depends on how general you want to get. If going from matrices to linear maps is general enough, then if you have $V$ and $W$ as finite-dimensional vector spaces over the same field $\mathbb{F}$, and if $A:V\rightarrow W$ is a linear map, we call the rank of $A$ the dimension of its range (which is as we know is a subspace of $W$), so $$ \mathrm{rank}A=\dim\mathrm{ran}A. $$
Now, what your statement says in the language of linear maps is that $$ \mathrm{rank}A=\mathrm{rank}A^*, $$ where $A^*$ is the adjoint or dual map of $A$ defined as $$(A^*\omega)(x)=\omega(Ax) $$ for any $\omega\in W^*$ and $x\in V$.
