Finding a matrix whose image matches kernel of a given matrix

Question

Given a matrix $A$, find another matrix $B$ such that $$\ker(A) = \operatorname{im}(B)$$ My approach:
If we are explicitly given the matrix(like all entries), then we can easily find a basis for the nullspace. Next, we can just represent B as a matrix with columns written as linear combination of the basis vectors of $ker(A)$. My doubt is can we do this in terms of some operation on A (and not explicitly find the nullspace basis)? Number of columns of B may be made anything in hopes of writing $B$ as some function $f(A)$
Extra: The root cause of my doubt arises from the 2 different ways of KVL,KCL given in the book 'Basic Circuit Theory' by Charles Desoer. The author writes KCL as $Aj=0$ and $v=A^Te$ where $A$ is the incidence matrix and $e$ is any vector
The second form is given under mesh analysis. $Mv=0$ and $M^Ti = j$ where $M$ is the mesh matrix and $i$ is any vector. $j$ and $v$ are the branch current and voltage respectively.

Answer 1

The kernel is space defined in an implicit way, i.e. by an equation (all $v$ such that $Av=0$), and the image of a transformation is defined in a explicit way, this space is parametrised. In general, I think there is no short cut, since getting an explicit representation involves solving a system of equations.

However, note that $$ \mathrm{ker}\,A = (\mathrm{im}\,A^T)^\perp, \tag{*}$$ so $\mathrm{im}\,A^T = (\mathrm{im}\, B)^\perp$. Maybe this explains why $A^T$ and $M^T$ occur in the KVL and KCL (I am not familiar with these formulations of the KL.)

Also in very specific cases, this relation above can help you. For instance, when $A$ is a regular $2\times 3$-matrix, then $B$ is the $3\times 1$-matrix $(a_1 \times a_2)^T$, where the $a_i$ are the rows of $A$. Geometrically: the kernel is a plane, and its normal vector determines the plane completely.

I hope this answer is useful to you.