Lets say, there is a transformation: T : <mi mathvariant="normal">&#x211C;<!-- ℜ -->

use the example in your comment: $M=\left(\begin{array}{ccc}4& 2& 1\\ 0& 1& 3\end{array}\right)$, with respect to the bases
$\mathrm{\beta <!--\; \beta \; -->}=\{v_1=[1,0,0],v_2=[1,1,0],v_3=[1,1,1]\}$
and
$\mathrm{\beta <!--\; \beta \; -->}{}^{\prime }=\{w_1=[1,0],w_2=[1,1]\}.$
This transformation takes as input a column vector
$v=\left(\begin{array}{c}{\mathrm{\alpha <!--\; \alpha \; -->}}_1\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_2\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_3\end{array}\right)$
representing the linear combination ${\mathrm{\alpha <!--\; \alpha \; -->}}_1{v}_1+{\mathrm{\alpha <!--\; \alpha \; -->}}_2{v}_2+{\mathrm{\alpha <!--\; \alpha \; -->}}_3{v}_3$ of basis vectors in $\mathrm{\beta <!--\; \beta \; -->}$ and produces as output the product
$Mv=\left(\begin{array}{ccc}4& 2& 1\\ 0& 1& 3\end{array}\right)\left(\begin{array}{c}{\mathrm{\alpha <!--\; \alpha \; -->}}_1\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_2\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_3\end{array}\right)=\left(\begin{array}{c}4{\mathrm{\alpha <!--\; \alpha \; -->}}_1+2{\mathrm{\alpha <!--\; \alpha \; -->}}_2+{\mathrm{\alpha <!--\; \alpha \; -->}}_3\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_2+3{\mathrm{\alpha <!--\; \alpha \; -->}}_3\end{array}\right).$
The entries $4{\mathrm{\alpha <!--\; \alpha \; -->}}_1+2{\mathrm{\alpha <!--\; \alpha \; -->}}_2+{\mathrm{\alpha <!--\; \alpha \; -->}}_3$ and ${\mathrm{\alpha <!--\; \alpha \; -->}}_2+3{\mathrm{\alpha <!--\; \alpha \; -->}}_3$ are then to be interpreted as coefficients of the vectors $w_1$ and $w_2$ in $\mathrm{\beta <!--\; \beta \; -->}{}^{\prime }$:
$T(v)=(4{\mathrm{\alpha <!--\; \alpha \; -->}}_1+2{\mathrm{\alpha <!--\; \alpha \; -->}}_2+{\mathrm{\alpha <!--\; \alpha \; -->}}_3)w_1+({\mathrm{\alpha <!--\; \alpha \; -->}}_2+3{\mathrm{\alpha <!--\; \alpha \; -->}}_3)w_2.$
If you want to know what this looks like in terms of the standard basis $\{[1,0],[0,1]\}$, just multiply out:
Note that this is exactly what you get from the product
$AMv=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{ccc}4& 2& 1\\ 0& 1& 3\end{array}\right)\left(\begin{array}{c}{\mathrm{\alpha <!--\; \alpha \; -->}}_1\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_2\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_3\end{array}\right),$
where $A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ is a change-of-basis matrix: it translates a representation in terms of $\mathrm{\beta <!--\; \beta \; -->}{}^{\prime }$ into one in terms of the standard basis. It’s easy to construct this change-of-basis matrix: its columns are just the representations of $w_1$ and $w_2$ in terms of the standard basis.
It follows that if you start with v, representing a vector in ${\mathbb{R}}^3$ in terms of the basis $\mathrm{\beta <!--\; \beta \; -->}$, and multiply it by the matrix
$AM=\left(\begin{array}{ccc}4& 3& 4\\ 0& 1& 3\end{array}\right),$
you get $T(v)$ expressed in terms of the standard basis for ${\mathbb{R}}^2$. Perhaps, though, you want to be able to input $v$ in terms of the standard basis for ${\mathbb{R}}^3$. Then you need another change-of-basis matrix, this time to convert from the standard basis for ${\mathbb{R}}^3$ to the basis $\mathrm{\beta <!--\; \beta \; -->}$ We already know how to go the other way: to transform from a representation in terms of $\mathrm{\beta <!--\; \beta \; -->}$ to one standard coordinates, multiply by the matrix
$B=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)$
whose columns are the representations of $w_1,w_2$ and $w_3$ in terms of the standard basis. (In other words, do exactly what we did to get $A$.) If you take the vector ${\mathrm{\alpha <!--\; \alpha \; -->}}_1{v}_1+{\mathrm{\alpha <!--\; \alpha \; -->}}_2{v}_2+{\mathrm{\alpha <!--\; \alpha \; -->}}_3{v}_3$ represented by the matrix
$v=\left(\begin{array}{c}{\mathrm{\alpha <!--\; \alpha \; -->}}_1\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_2\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_3\end{array}\right)$
in terms of the basis $\mathrm{\beta <!--\; \beta \; -->}$ , you can find its representation in terms of the standard basis by multiplying by $B$ to get
$Bv=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}{\mathrm{\alpha <!--\; \alpha \; -->}}_1\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_2\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_3\end{array}\right)=\left(\begin{array}{c}{\mathrm{\alpha <!--\; \alpha \; -->}}_1+{\mathrm{\alpha <!--\; \alpha \; -->}}_2+{\mathrm{\alpha <!--\; \alpha \; -->}}_3\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_2+{\mathrm{\alpha <!--\; \alpha \; -->}}_3\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_3\end{array}\right).$
Unfortunately, this isn’t quite what we need: we want to start with a vector $v$ in standard coordinates and convert it to one in $\mathrm{\beta <!--\; \beta \; -->}$ coordinates so that we can multiply by $AM$ and get $T(v)$ in standard coordinates. That requires changing base from standard to $\mathrm{\beta <!--\; \beta \; -->}$; multiplying by $B$ goes in the opposite direction, from $\mathrm{\beta <!--\; \beta \; -->}$ coordinates to standard ones. As you might expect, the matrix that does the change of basis in the other direction is $B^{-<!--\; -\; -->1}$, which I’ll let you compute for yourself. Once you have it, you can express $T$ in terms of a matrix multiplication that involves standard coordinates on both ends:
$T(v)=AM{B}^{-<!--\; -\; -->1}v$
gives $T(v)$ in standard ${\mathbb{R}}^2$ coordinates if $v$ is expressed in standard ${\mathbb{R}}^3$ coordinates.