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Lets say, there is a transformation: $T:\Re ^{n}\rightarrow \Re ^{m}$ transforming a vector in $V$ to $W$. Now the transformation matrix, $A=\begin{bmatrix} a_{11} & a_{12} &...&a_{1n} \\ a_{21} & a_{22} &...&a_{2n} \\ . & . & .\\ . & . & .\\ . & . & .\\ a_{n1} &a_{n2} &...&a_{n3}\\ \end{bmatrix}$

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 $T$

Soham

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Do you mean that you want to find $T$ in terms of the standard basis vectors? –  Brian M. Scott Jun 15 '12 at 7:20
    
@Brian I want to find T given a transofrmation matrix A, and the basis vectors of V and W. The basis vectors of V and W are not standard, and I am intending to understand the process and then perhaps somewhere I will "constrain" the basis vectors to be standard. But now, I am not making any assumption, and want to generalize the "algorithm" if I can say so. –  Soham Jun 15 '12 at 7:25
    
I don’t understand: you have $T$, because $A$ contains all the information that there is about $T$. –  Brian M. Scott Jun 15 '12 at 7:27
    
Let M = $M=\begin{pmatrix} 4 & 2 & 1\\ 0 & 1 & 3 \end{pmatrix}$ and $\beta = (v_{1}=[1,0,0],v_{2}=[1,1,0],v_3=[1,1,1])$ as its basis vectors and $\beta ' =(w_1=[1,0],w_2=[1,1])$ then find the unique transformation $T$ such that $T:\mathbb{R}^{n} \to \mathbb{R}^{m}$ This got me thinking that though I am able to easily find $A$ (in T(x)=Ax) when T is given, I am not able to find T when there are non standard vectors involved. [Or do you suggest I decompose v2 and v3 in standard vectors] –  Soham Jun 15 '12 at 7:42

1 Answer 1

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I’ll use the example in your comment: $M=\pmatrix{4&2&1\\0&1&3}$, with respect to the bases $$\beta=\{v_1=[1,0,0],v_2=[1,1,0],v_3=[1,1,1]\}$$ and $$\beta\,'=\{w_1=[1,0],w_2=[1,1]\}\;.$$

This transformation takes as input a column vector $$v=\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}$$

representing the linear combination $\alpha_1v_1+\alpha_2v_2+\alpha_3v_3$ of basis vectors in $\beta$ and produces as output the product

$$Mv=\pmatrix{4&2&1\\0&1&3}\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}=\pmatrix{4\alpha_1+2\alpha_2+\alpha_3\\\alpha_2+3\alpha_3}\;.$$

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 $\beta\,'$:

$$T(v)=(4\alpha_1+2\alpha_2+\alpha_3)w_1+(\alpha_2+3\alpha_3)w_2\;.$$

If you want to know what this looks like in terms of the standard basis $\{[1,0],[0,1]\}$, just multiply out:

$$\begin{align*} T(v)&=(4\alpha_1+2\alpha_2+\alpha_3)\pmatrix{1\\0}+(\alpha_2+3\alpha_3)\pmatrix{1\\1}\\ &=\pmatrix{4\alpha_1+3\alpha_2+4\alpha_3\\\alpha_2+3\alpha_3}\;. \end{align*}$$

Note that this is exactly what you get from the product

$$AMv=\pmatrix{1&1\\0&1}\pmatrix{4&2&1\\0&1&3}\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}\;,$$

where $A=\pmatrix{1&1\\0&1}$ is a change-of-basis matrix: it translates a representation in terms of $\beta\,'$ 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 $\Bbb R^3$ in terms of the basis $\beta$, and multiply it by the matrix $$AM=\pmatrix{4&3&4\\0&1&3}\;,$$ you get $T(v)$ expressed in terms of the standard basis for $\Bbb R^2$. Perhaps, though, you want to be able to input $v$ in terms of the standard basis for $\Bbb R^3$. Then you need another change-of-basis matrix, this time to convert from the standard basis for $\Bbb R^3$ 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 $$B=\pmatrix{1&1&1\\0&1&1\\0&0&1}$$ 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 $\alpha_1v_1+\alpha_2v_2+\alpha_3v_3$, represented by the matrix $$v=\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}$$ in terms of the basis $\beta$, you can find its representation in terms of the standard basis by multiplying by $B$ to get $$Bv=\pmatrix{1&1&1\\0&1&1\\0&0&1}\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}=\pmatrix{\alpha_1+\alpha_2+\alpha_3\\\alpha_2+\alpha_3\\\alpha_3}\;.$$

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:

$$T(v)=AMB^{-1}v$$

gives $T(v)$ in standard $\Bbb R^2$ coordinates if $v$ is expressed in standard $\Bbb R^3$ coordinates.

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wow! Thanks so much, not only for the solution but the entire explanation and the care you took in forming the answer. Just wow! I did a bit of reading before catching upon your answer and I realized how shaky my concepts were. Soham –  Soham Jun 15 '12 at 16:25

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