I am trying to understand this question here, but I'm getting lost figuring out how the map is a homomorphism. Could someone please explain it to me? The original question was:
Let $F$ be a field. Given an $(m \times n)-$matrix $$ M = (t_{ij})_{1 \leq i \leq m,\, 1 \leq j \leq n}$$
over $F$, define $rank(M)$ as the dimension of the subspace $Col(M)$ of $F^{m}$ spanned by all columns $(t_{ip})_{1 \leq i \leq m,\, p=1,2,\cdots , n}$.
Consider the set $S \subseteq F^{n}$ of all solutions $\{x_{1},x_{2},\cdots , x_{n}\}$ of the homogeneous linear system of equations over $F$:
$$(15.3) \begin{cases} t_{11}x_{1}+t_{12}x_{2}+\cdots + t_{1n}x_{n}=0, \\ t_{21}x_{1}+t_{22}x_{2}+\cdots + t_{2n}x_{n} = 0, \\ \vdots \\ t_{m1}x_{1}+t_{m2}x_{2} + \cdots + t_{mn}x_{n} = 0 \end{cases} $$
I need to find a homomorphism $f: F^{n} \to F^{m}$ such that $S = \ker(f)$.
I now that usually, for $M \in M_{mn}$, the map $T:x \mapsto Mx$ does the job of having the solution set equal to the kernel. Does that work here? It's a linear transformation and not a "homomorphism" per se, but I've found that they work pretty much the same way.
Aside from this, assuming what I've said here is correct, is there any kind of formal proving process I need to go through, and if so, could you give me any pointers to that effect?
I understand why the solution set is the kernel, but I'm not able to show its a homomorphism. Is it because you can consider the inverse map $T^{-1}: y = Mx \rightarrow M^{-1}y$, which makes T a bijection and then you can consider elements $x,y$ and then $T(x+y) = M(x+y) = Mx + My = T(x) + T(y)$.
The comment in the previous post that is confusing me the most is "You only have to associtae a linear mmap to the matrix. This simply means choosing a basis, e.g. the canonical basis." Could someone please explain what this means/how to do it?