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Given finite dimensional vector spaces $V$ and $W$, suppose we have a linear transformation $T:V \rightarrow W$ and an arbitrary basis $B = \{ b_1,b_2,\ldots,b_n\}$ in the $n$-dimensional vector space $V$.
Since $T$ is linear and all elements $x \in V$ can be represented as $x = c_1b_1 + \ldots +c_nb_n$,
it's clear that $$T(x) = c_1T(b_1) + \ldots + c_nT(b_n) = A [x]B$$ where $A$ has the transformed column vectors of $B$ .

1 - Does that mean the matrix representation of ANY linear transformation starting from $V$ is applied ( multiplies ) not to a vector from $V$, but actually to a coordinate vector of $x$ ( contained in $\mathbb{R}^n$, isomorphic to $V$ ) with respect to some basis ?

2 -Are the column vectors of A actually T(b1), T(b2),..,T(bn) or the coordinate vectors of T(b1),T(b2),...,T(bn) w.r.t to some basis D={d1,d2,...,dk} in W ?

If the latter is positive, then will the result of A [x]B actually yield not T(x) but the coordinate vectors of T(x) w.r.t to the basis D in W ?

But then why $$T(x) = c_1T(b_1) + \ldots + c_nT(b_n) = A [x]B$$ tells me that the matrix should have vectors T(b1),T(b2),..,T(bn) and not coordinate vectors of T(b1),T(b2),...,T(bn) w.r.t to some basis D in W ? And why does the result of the matrix multiplication results in T(x), and not a coordinate vector of T(x) w.r.t to some basis D in W ?

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Please check that my edit did not destroy your message. I'm not sure I got the centred equation right.. –  Mårten W Apr 21 '13 at 8:52
    
Thanks for the edit, i don't know how to use laTex. –  nerdy Apr 21 '13 at 8:58
    
I thought I edited point 2 as well, but it does not look edited.. –  Mårten W Apr 21 '13 at 9:01
    
I changed it because i clarified a bit my doubt. –  nerdy Apr 21 '13 at 9:06
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2 Answers

Yes and yes. Matrices can only be applied to coordinate vectors, columns formed of numbers. You cannot apply a matrix to an abstract vector. And in order to have a matrix for a linear transformation, you need bases both at departure and at arrival; the columns of $A$ are the images by $T$ of the chosen basis vectors in $V$ expressed in the chosen basis of $W$.

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But then why the equation T(x)=c1T(b1)+…+cnT(bn)=A[x]B tells me that the matrix should have abstract vectors T(b1),T(b2),..,T(bn) as its columns and not coordinate vectors of T(b1),T(b2),...,T(bn) w.r.t to some basis D in W ? And why does the same equation shows that the matrix multiplication results in an abstract vector T(x), and not a coordinate vector of T(x) w.r.t to some basis D in W ? –  nerdy Apr 21 '13 at 9:16
    
I cannot make any sense out of $A[x]B$; your matrix should not depend on $x$ (or did you mean a cross?); and in any case you don't have any matrix until you have chosen a basis in $W$. Some people might entertain in writing matrices with non-numeric entries, in which case you could write $c_1T(b_1)+…+c_nT(b_n)$ as a $(1\times n)\times(n\times 1)$ product $X\cdot C$ where $X=(T(B_1)~\cdots~T(B_n))$ is a $1\times n$ matrix with abstract vector entries $T(B_i)$, and $C$ is a column of the numbers $c_i$, but I don't think this point of view clarifies anything. –  Marc van Leeuwen Apr 21 '13 at 9:31
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Too long for a comment:

To get a matrix rep. for $\,T\,$ you also must choose a basis in the codomain $\,W\,$, which you didn't do. What you wrote seems to imply that a basis you choose is formed with the vectors $\,\{Tb_1,\ldots,Tb_n\}\,$ , but this may be pretty problematic as this set may easily be linearly dependent, so you'll have to take out the linearly dependent part...and even then, when you're left with a linearly independent set, you must probably have a lot of ways to complete it to a basis of $\,W\,$ ...

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