Relation between left null space, row space and cokernel, coimage I know the null space and the column space of a matrix $A$ can be identified with the kernel and the image of the associated homomorphism $L_A: x \mapsto Ax$, once a basis is chosen.
I was reading this Wikipedia article that calls the left null space and the row space of $A$ respectively cokernel and coimage. This seems to imply that (using my notation) $\text{coker } L_A = \ker L^t_A = \ker L_{A^t}$ which in turn implies $\text{coker }A$ is the orthogonal complement of $\text{im } A$.
However the cokernel of the homomorphism $L: V \rightarrow W$ is defined here as $W/\text{im L}$, the set of all equivalence classes of every vector of $\text{im } L$. This means $\text{coker }L$ and $\ker L^t$ have different types entirely and that relation can't possibly be true.
Is this true maybe in some other sense? What is the actual relation?

Suppose the matrix $A$ is $m\times n$ and has rank $k$.
Thus $A$ defines a linear map $L_A\colon \mathbb{R}^n\to \mathbb{R}^m$. The null space has dimension $n-k$ and the image has dimension $k$.
The cokernel $\mathbb{R}^{m}/\operatorname{im}L_A$ has dimension $m-k$. The coimage $\mathbb{R}^n/\ker L_A$ has dimension $n-(n-k)=k$.
The matrix $A$ also defines a linear map $R_A\colon \mathbb{R}_m\to \mathbb{R}_n$ (spaces of row vectors) by $y\mapsto yA$. The image of this linear map has dimension $k$ and the kernel has dimension $m-k$.
Hey! The dimensions agree! Maybe we can find a “canonical” isomorphism
$$
f\colon \ker R_A\to \operatorname{coker}L_A
$$
In other words, we'd like to associate to each vector $y\in\ker R_A$ a unique element in $F^m/\operatorname{im}L_A$.
Let $y\in\ker R_A$, so $y^T\in \mathbb{R}^m$. The natural choice would be defining $f(y)=y^T+\operatorname{im}L_A$.
This is clearly a linear map. Is it injective? Suppose $y\in\ker f$. This means $y^T=Ax$ for some $x\in F^n$. Also $yA=0$ by definition, so
$$
0=A^Ty^T=A^TAx
$$
and so $x^TA^TAx=0$ and therefore $x=0$; hence $y^T=Ax=0$. Yes! The map $f$ is injective. The dimensions coincide, so $f$ is indeed an isomorphism.
Similarly you can find an isomorphism between $\mathbb{R}^n/\ker L_A$ and $\operatorname{im}R_A$.
Note however that these isomorphisms require that $x^TA^TAx=0$ implies $x=0$. This is not true over general fields.
