I am considering the mapping $ u : \mathcal{M}_{\mathbb{K}}(n,p)\to L(\mathbb{K}^p,\mathbb{K}^n) : M\mapsto l_{M} $ and I would like to show this is a vector space isomorphism.
I should mention that $l_{M}(X) = MX$ where $X\in\mathbb{K}^p$ and $M\in\mathcal{M}_{\mathbb{K}}(n,p)$
First, I want to show that :
$\forall A,B\in\mathcal{M}_{\mathbb{K}}(n,p)\times\mathcal{M}_{\mathbb{K}}(n,p),\forall\alpha\in\mathbb{R} : u(\alpha A+ B) = \alpha u(A) + u(B)$
Since $u(A)$ and $u(B)$ are functions of elements in $\mathbb{K}^p$, the previous equality needs to hold for all $X\in\mathbb{K}^p$.
In fact, the set $L(\mathbb{K}^p,\mathbb{K}^n)$ is a vector space for the addition of vector and multiplication by a scalar that is defined on the image of the mapping of this set.
But for all $X\in\mathbb{K}^p$ we have :
$u(\alpha A + B)(X) = (\alpha A + B)X = \alpha AX + BX = \alpha u(A)(X) + u(B)(X)$
So it is a linear mapping. Now we need to show that $u$ is bijective.
Consider $Z\in Ker(u) = \left\{ X\in\mathcal{M}_{\mathbb{K}}(n,p) : u(X) = 0_{m} \right\}$ where $0_{m}\in L(\mathbb{K}^p,\mathbb{K}^n)$ is the null mapping. It means that :
$\forall X\in\mathbb{K}^p : u(Z)(X) = ZX = 0_{n}\implies Z = 0_{n,p} $ which shows that $u$ is injective.
Now, to show that it is bijective, I don't have any idea except the one consisting to show that the rank of $u$ is equal to the dimension of $L(\mathbb{K}^p,\mathbb{K}^n)$ but I don't know how to proceed ?
Thank you a lot !
What you've done looks good. As you said it remains to show that the mapping is surjective.
Therefore let $l : \mathbb K^p \rightarrow \mathbb K^n$ be a linear map. By considering the canonical basis of $\mathbb K^p$ and $\mathbb K^m$ then you can find a matrix $M_l$ such that for all $X$ you get
$$ M_lX = l(X).$$ That is, you can find $M_l$ such that $u(M_l) = l.$