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December 15th, 2017, 09:55 AM   #1
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Linear application and matrix.

Hello everyone, I have a doubt. I know that the image of a generic domain vector, defined by the linear application, is equivalent to the vector-matrix product between the matrix associated with the application and a generic domain vector. Is the associated matrix that we use always built with canonical bases or can you use associated arrays built with a generic base of the domain and a generic base of the codomain? Thank you all
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December 15th, 2017, 11:32 AM   #2
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Given any basis, not just "canonical" bases, there exist a matrix representing the linear transformation in that basis. In fact, you can have one basis for the "domain" space and another for the range space.

For example, suppose T is the linear transformation that maps (x, y, z) into (3x+ y, y- z, x+ y+ z). Take the basis for the domain space to be {(1, 0, 1), (1, 1, 0), (0, 1, 1)} and the basis for the range space to be {(1, 0, 0), (1, 1, 1), (3, 2, 1)}.

Apply T to (1, 0, 1). That gives (3+ 0,0- 1, 1+ 0+ 1)= (3, -1, 2). Write that as a linear combination of {(1, 0, 0), (1, 1, 1), (3, 2, 1)}: (3, -1, 2)= a(1, 0, 0)+ b(1, 1, 1)+ c(3, 2, 1)= (a+ b+ 3c, b+ 2c, b+ c). That gives the three equations a+ b+ 3c= 3, b+ 2c= -1, and b+ c= 2. Subtracting the last equation from the second, c= -3. Then b+ (-3)= 2 so b= 5. a+ b+ 3c= a+ 5- 3= a+ 2= 3 so a= 1. The first column of the matrix representation of T, in those bases, is $\displaystyle \begin{pmatrix}1 \\ 5 \\ -3\end{pmatrix}$.

Apply T to (1, 1, 0) and write the result as a linear combination of {(1, 0, 0), (1, 1, 1), (3, 2, 1)} to get the second column and to (0, 1, 1) to get the third column.
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December 15th, 2017, 12:31 PM   #3
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Hello Country Boy, yes i know how to built a matrix associated to a linear application.

I mean this:
T is a linear application
T: R^2->R^3

1 2 is the matrix associated to this application
4 7
7 8

(canonical bases)


I can calculate the image of a generic vector of the domain through the vector product matrix between the representative matrix and a generic domain vector.


1 2
4 7 (x,y) = (x+2y,4x+5y,7x+8y)
7 8

For example

T(2,1) = (4,13,24)

But if I make this product with a representative matrix constructed from non-canonical bases then the image of this same vector will be different from the previous one.

So I do not know if I should report (to define the image of a generic vector of the domain) only to matrixes built in relation to the canonical bases. Thank you very much.Sorry for my english.

Last edited by JackPirri; December 15th, 2017 at 12:36 PM.
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