Is the following Proof Correct? In addition can the argument expressed more eloquently.
Proposition. Let $M$ and $A$ be matrices such that the product $MA$ is defined. Prove that if the $j^{th}$ column of $A$ is a linear combination of a set of columns of $A$ then the $j^{th}$ column of $MA$ is a linear combination of the corresponding columns of $A$ with the same coefficients.
Proof. Let $M$ by an $p\times n$ and $A$ be an $n\times q$ matrix. Then we can be assured that the product $MA$ is defined. Now assume that for some set $\beta = \{x_1,x_2,...,x_r\}\subseteq\{1,2,...,q\}$ the $j^{th}$ column of $A$ is such that $$\begin{pmatrix}A_{1j}\\A_{2j}\\.\\.\\.\\A_{nj}\end{pmatrix} = C_1\begin{pmatrix}A_{1x_1}\\A_{2x_2}\\.\\.\\.\\A_{nx_2}\end{pmatrix}+ C_2\begin{pmatrix}A_{1x_2}\\A_{2x_2}\\.\\.\\.\\A_{nx_2}\end{pmatrix}+ \cdot\cdot\cdot+ C_r\begin{pmatrix}A_{1x_r}\\A_{2x_r}\\.\\.\\.\\A_{nx_r}\end{pmatrix}$$
Then by making use of the distributivity of matrix multiplication. The $j^{th}$ column of $MA$ can be computed as follows
$$M\cdot\begin{pmatrix}A_{1j}\\A_{2j}\\.\\.\\.\\A_{nj}\end{pmatrix} = C_1\cdot M\begin{pmatrix}A_{1x_1}\\A_{2x_2}\\.\\.\\.\\A_{nx_2}\end{pmatrix}+ C_2\cdot M\begin{pmatrix}A_{1x_2}\\A_{2x_2}\\.\\.\\.\\A_{nx_2}\end{pmatrix}+ \cdot\cdot\cdot+ C_r\cdot M\begin{pmatrix}A_{1x_r}\\A_{2x_r}\\.\\.\\.\\A_{nx_r}\end{pmatrix}$$
Lastly using the definition of matrix multiplication we recognize the products of the form $$M\begin{pmatrix}A_{1x_1}\\A_{2x_2}\\.\\.\\.\\A_{nx_2}\end{pmatrix}$$ to be the $x_i^{th}$ columns of the product $MA$ this observation together with the preceeding equation then yields the require result.
$\blacksquare$
Your proof seems fine to me.
An alternative point of view would be that of linear algebra: let $A$ stand for a set of $q$ vectors of $\mathbb{K}^n$ (its columns) and $M$ be the matrix of a linear application from $\mathbb{K}^n$ to $\mathbb{K}^p$. Your condition on $A$ means that the set of its columns is linearly dependent. What you are then showing is that the image of that set when $M$ is applied is still a linearly dependent collection in $\mathbb{K}^p$. You are even showing something a bit stronger: the coefficients remain the same (which is quite obvious since $M$ is a linear transformation).