Apparently a Theorem exists that says: the rank of a linear map coincides with the rank of any of its matrix representations.
Context is from this question Making matrices look the same w.r.t different basis's
i am looking for said theorem and preferably a proof of it, i tried to prove the result on my own but im apparently to braindead atm to do so.
Suppose $\varphi: V\to W$ is a linear map, where $V$ and $W$ are finite-dimensional $\mathbb{F}$-vector spaces. Suppose $(\mathbf{v}_1,\ldots,\mathbf{v}_n)$ and $(\mathbf{w}_1,\ldots,\mathbf{w}_m)$ are ordered bases for $V$ and $W$, respectively, giving rise to a matrix representation $\mathbf{A} \in \mathbb{F}^{m\times n}$ of $\varphi$.
By definition, $\operatorname{rank}(\mathbf{A})$ is the column rank of $\mathbf{A}$, i.e. the dimension of the linear span of the columns of $\mathbf{A}$, call them $\mathbf{a}_1,\ldots,\mathbf{a}_n$. But $$\mathbf{A}\mathbf{v} = \mathbf{a}_1 \mathbf{v}(1) + \cdots + \mathbf{a}_n \mathbf{v}(n)$$ shows that in fact the linear span of $\{\mathbf{a}_1,\ldots,\mathbf{a}_n\}$ is the image of $\mathbb{F}^n \cong V$ under the map $\mathbf{A}$ (i.e. multiplication by $\mathbf{A}$ on the left).
The isomorphism $W \cong \mathbb{F}^m$ given by the ordered basis above provides, upon restriction, an isomorphism between the image of $\mathbf{A}$ (as a subspace of $\mathbb{F}^m$) and the image of $\varphi$ (as a subspace of $W$).
In particular, the dimension of the image of $\mathbf{A}$ and the dimension of the image of $\varphi$ are exactly the same. But the former was $\operatorname{rank}(\mathbf{A})$, so $$\operatorname{rank}(\mathbf{A}) = \dim \varphi[V]$$ This was independent of the choice of ordered bases for $V$ and $W$, so it follows that $$\operatorname{rank}(\mathbf{A}) = \dim \varphi[V] = \operatorname{rank}(\mathbf{B})$$ for any matrix representations $\mathbf{A},\mathbf{B}$ of $\varphi$.