 # Lets say, there is a transformation: T : <mi mathvariant="normal">&#x211C;<!-- ℜ --> sebadillab0 2022-07-07 Answered
Lets say, there is a transformation: $T:{\mathrm{\Re }}^{n}\to {\mathrm{\Re }}^{m}$ transforming a vector in $V$ to $W$. Now the transformation matrix,
$A=\left[\begin{array}{cccc}{a}_{11}& {a}_{12}& ...& {a}_{1n}\\ {a}_{21}& {a}_{22}& ...& {a}_{2n}\\ .& .& .\\ .& .& .\\ .& .& .\\ {a}_{n1}& {a}_{n2}& ...& {a}_{n3}\end{array}\right]$
The basis vectors of $V$ are ${v}_{1},{v}_{2},{v}_{3}...,{v}_{n}$ which are all non standard vectors and similarly ${w}_{1},{w}_{2},...,{w}_{m}$
My question is, in the absence of the basis vectors being standard vectors what is the procedure of finding $T$
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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
$\beta =\left\{{v}_{1}=\left[1,0,0\right],{v}_{2}=\left[1,1,0\right],{v}_{3}=\left[1,1,1\right]\right\}$
and
$\beta {\phantom{\rule{thinmathspace}{0ex}}}^{\prime }=\left\{{w}_{1}=\left[1,0\right],{w}_{2}=\left[1,1\right]\right\}\phantom{\rule{thickmathspace}{0ex}}.$
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 $\beta {\phantom{\rule{thinmathspace}{0ex}}}^{\prime }$:
$T\left(v\right)=\left(4{\alpha }_{1}+2{\alpha }_{2}+{\alpha }_{3}\right){w}_{1}+\left({\alpha }_{2}+3{\alpha }_{3}\right){w}_{2}\phantom{\rule{thickmathspace}{0ex}}.$
If you want to know what this looks like in terms of the standard basis $\left\{\left[1,0\right],\left[0,1\right]\right\}$, 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 $T\left(v\right)$ 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 $T\left(v\right)$ 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:
$T\left(v\right)=AM{B}^{-1}v$
gives $T\left(v\right)$ in standard ${\mathbb{R}}^{2}$ coordinates if $v$ is expressed in standard ${\mathbb{R}}^{3}$ coordinates.

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