Question: A Linear transformation $T: \mathbb R^4 \to \mathbb R^4$ is represented by the matrix $\mathbf A=\begin{pmatrix} \\1&-1&2&3 \\ 2 & -3 & 4 & 5\\ 5 & -6 & 10 & 14\\ 4 & -5 & 8 & 11 \end{pmatrix}$ (i) Show that the $\dim\big(R(\mathbf A)\big) =2$.
(ii) Let M be a given $4\times 4$ (non-singular) matrix and let $\mathcal S$ be the vector space consisting of vectors in the form of MAx where $x \in \mathbb R^4$. Show that if M is non-singular then $\dim\big(\mathcal S\big) =2$.
The first part is pretty straight-forward. All one needs to do is compute $\operatorname{rref}(\mathbf A)$.
But it is the second part that has stomped me (and for quite some time now). I did come up with a solution but it seemed very cheap to me and it didn't really satisfy me completely.
What I tried was to write A in the form, $\mathbf A=\Big [\begin{matrix} \vec p_1& \vec p_2&\vec f_1&\vec f_2 \\ \end{matrix}\Big]$ Where $\vec p_1$ and $\vec p_2$ are the columns corresponding to the pivot variables (there are $2$ pivot variables, since $\dim\big(R(\mathbf A)\big) =2$) and $ \vec f_1$ and $\vec f_2$ are the columns corresponding to the free variables.
Now if M be any $4\times 4$ matrix in the form, $\mathbf M=\Big[ \begin{matrix} \vec c_1& \vec c_2& \vec c_3& \vec c_4 \\ \end{matrix}\Big]$ Where $\vec c_1, \vec c_2, \vec c_3$ and $\vec c_4$ are the four columns of M, then $\mathbf {MAx}=\Big[ \begin{matrix} \vec c_1& \vec c_2& \vec c_3& \vec c_4 \\ \end{matrix}\Big] \cdot x_1\vec p_1+x_2\vec p_2+ x_3\vec f_1+x_4\vec f_2 $ I don't know what to do after this.
Is this apprach correct? If yes, how do I proceed from here? If not, where is it that I am going wrong and how do I solve this problem?
Any Help is much appreciated!
Thanks in Advance!