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

$\beta =\{{v}_{1}=[1,0,0],{v}_{2}=[1,1,0],{v}_{3}=[1,1,1]\}$

and

$\beta {\phantom{\rule{thinmathspace}{0ex}}}^{\prime}=\{{w}_{1}=[1,0],{w}_{2}=[1,1]\}\phantom{\rule{thickmathspace}{0ex}}.$

This transformation takes as input a column vector

$v=\left(\begin{array}{c}{\alpha}_{1}\\ {\alpha}_{2}\\ {\alpha}_{3}\end{array}\right)$

representing the linear combination ${\alpha}_{1}{v}_{1}+{\alpha}_{2}{v}_{2}+{\alpha}_{3}{v}_{3}$ of basis vectors in $\beta $ and produces as output the product

$Mv=\left(\begin{array}{ccc}4& 2& 1\\ 0& 1& 3\end{array}\right)\left(\begin{array}{c}{\alpha}_{1}\\ {\alpha}_{2}\\ {\alpha}_{3}\end{array}\right)=\left(\begin{array}{c}4{\alpha}_{1}+2{\alpha}_{2}+{\alpha}_{3}\\ {\alpha}_{2}+3{\alpha}_{3}\end{array}\right)\phantom{\rule{thickmathspace}{0ex}}.$

The entries $4{\alpha}_{1}+2{\alpha}_{2}+{\alpha}_{3}$ and ${\alpha}_{2}+3{\alpha}_{3}$ are then to be interpreted as coefficients of the vectors ${w}_{1}$ and ${w}_{2}$ in $\beta {\phantom{\rule{thinmathspace}{0ex}}}^{\prime}$:

$T(v)=(4{\alpha}_{1}+2{\alpha}_{2}+{\alpha}_{3}){w}_{1}+({\alpha}_{2}+3{\alpha}_{3}){w}_{2}\phantom{\rule{thickmathspace}{0ex}}.$

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

$AMv=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{ccc}4& 2& 1\\ 0& 1& 3\end{array}\right)\left(\begin{array}{c}{\alpha}_{1}\\ {\alpha}_{2}\\ {\alpha}_{3}\end{array}\right)\phantom{\rule{thickmathspace}{0ex}},$

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 $\beta {\phantom{\rule{thinmathspace}{0ex}}}^{\prime}$ 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 $\beta $, and multiply it by the matrix

$AM=\left(\begin{array}{ccc}4& 3& 4\\ 0& 1& 3\end{array}\right)\phantom{\rule{thickmathspace}{0ex}},$

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 $\beta $ We already know how to go the other way: to transform from a representation in terms of $\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 ${\alpha}_{1}{v}_{1}+{\alpha}_{2}{v}_{2}+{\alpha}_{3}{v}_{3}$ represented by the matrix

$v=\left(\begin{array}{c}{\alpha}_{1}\\ {\alpha}_{2}\\ {\alpha}_{3}\end{array}\right)$

in terms of the basis $\beta $ , you can find its representation in terms of the standard basis by multiplying by $B$ to get

$Bv=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}{\alpha}_{1}\\ {\alpha}_{2}\\ {\alpha}_{3}\end{array}\right)=\left(\begin{array}{c}{\alpha}_{1}+{\alpha}_{2}+{\alpha}_{3}\\ {\alpha}_{2}+{\alpha}_{3}\\ {\alpha}_{3}\end{array}\right)\phantom{\rule{thickmathspace}{0ex}}.$

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 $\beta $ coordinates so that we can multiply by $AM$ and get $T(v)$ in standard coordinates. That requires changing base from standard to $\beta $; multiplying by $B$ goes in the opposite direction, from $\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.