I Linear transformations

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Given w = T (v), where T is a linear transformation and w and v are vectors, why is it that we can write any coefficient of w, such as w1 as a linear combination of the coefficients of v? i.e. w1 = av1 + bv2 + cv3

Supposably this is a consequence of the definition of linear transformations, but I don't see the connection.
 

fresh_42

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In order to write ##w=(w_1,w_2,w_3)## and ##v=(v_1,v_2,v_3)## what do you need? And what does this mean for ##T##?
 
In order to write ##w=(w_1,w_2,w_3)## and ##v=(v_1,v_2,v_3)## what do you need? And what does this mean for ##T##?
We need a basis to be able to express a vector in terms of coefficents. I don't know what your second question implies.
 

fresh_42

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##T## is a linear transformation. What does this mean? Sure, it means that ##T(\alpha u + \beta v)=\alpha T(u)+\beta T(v)##, but since we have a basis, we can express this transformation in terms of the basis, too:
We write down ##T(1,0,\ldots ,0)\, , \, \ldots \, , \,T(0,\ldots,0,1)## as column vectors with respect to the same basis we used to write ##w##. The result is a number scheme which we call a matrix. Now it turns out that ##w=Tv## is exactly the matrix multiplication ##(w_1,\ldots,w_m) = \left( \sum_{j=1}^n T_{ij}v_j \right)_{1\leq i \leq m}## where ##n=\dim V## and ##m=\dim W##.

So the coefficients of ##v## and ##w## need a bases to write them as a tuple of components, and the same bases allow us to write the linear transformation as a number scheme, too.

Your example had ##n=m=3## and the coefficients are ##a=T_{11}, b=T_{12},c=T_{13}##.
 
##T## is a linear transformation. What does this mean? Sure, it means that ##T(\alpha u + \beta v)=\alpha T(u)+\beta T(v)##, but since we have a basis, we can express this transformation in terms of the basis, too:
We write down ##T(1,0,\ldots ,0)\, , \, \ldots \, , \,T(0,\ldots,0,1)## as column vectors with respect to the same basis we used to write ##w##. The result is a number scheme which we call a matrix. Now it turns out that ##w=Tv## is exactly the matrix multiplication ##(w_1,\ldots,w_m) = \left( \sum_{j=1}^n T_{ij}v_j \right)_{1\leq i \leq m}## where ##n=\dim V## and ##m=\dim W##.

So the coefficients of ##v## and ##w## need a bases to write them as a tuple of components, and the same bases allow us to write the linear transformation as a number scheme, too.

Your example had ##n=m=3## and the coefficients are ##a=T_{11}, b=T_{12},c=T_{13}##.
 

WWGD

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Given ##T: V \rightarrow W ##, Consider a basis ## \{ v_1, v_2,...,v_n\} ## for ##V ##. Then ##T(v)=.... ##
 
##T## is a linear transformation. What does this mean? Sure, it means that ##T(\alpha u + \beta v)=\alpha T(u)+\beta T(v)##, but since we have a basis, we can express this transformation in terms of the basis, too:
We write down ##T(1,0,\ldots ,0)\, , \, \ldots \, , \,T(0,\ldots,0,1)## as column vectors with respect to the same basis we used to write ##w##. The result is a number scheme which we call a matrix. Now it turns out that ##w=Tv## is exactly the matrix multiplication ##(w_1,\ldots,w_m) = \left( \sum_{j=1}^n T_{ij}v_j \right)_{1\leq i \leq m}## where ##n=\dim V## and ##m=\dim W##.

So the coefficients of ##v## and ##w## need a bases to write them as a tuple of components, and the same bases allow us to write the linear transformation as a number scheme, too.

Your example had ##n=m=3## and the coefficients are ##a=T_{11}, b=T_{12},c=T_{13}##.
Got it!
 

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