Not clear about the change of basis in new space

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Actually after I wrote down the query on the invertible matrix which I posted a few days ago I happened to refer again to Kunze Huffman and found that this is a standard theorem regarding transformation of linear operator from one basis to another.

Then I realized that the point which was not clear was that if Tv is a vector in basis B then how could with respect to B' I could write P(Tv) where P is the matrix of transformation from B to B'. What is unclear is that when u are doing this u are actually trying to premultiply a vector which is already in the space W, but according to theorem 8 on pg 53 of the book it says that

Suppose I is an n x n invertible matrix over F. Let V be an n dimensional vector space over F and let B be an ordered basis of V . Then there exists unique ordered basis B' of V such that [alpha] in basis B= P[alpha ] in basis B'.

So how is it here the vector space in which the vector is going to reside and the basis are completely different. Am I missing something very obvious. My thinking is that probably even if the space W has smaller dimension than V it is extended by adding 0s to it to equate V and then trying to apply the above technique. Still I am highly confused of applying a matrix NxN which would transform V -> V on something in W.

Sorry but I do not know how to use the subscripts here for clarity but hopefully I have been able to make my doubt across.
 

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  • #2
siddharth
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So how is it here the vector space in which the vector is going to reside and the basis are completely different.
It's possible for a vector space to have an infinite number of basis. For example, take the vector space [itex] \mathbb{R}^2[/itex]. Now [itex]B_1=\{ (0,1) , (1,0) \} [/itex] is a basis for the vector space. Notice that [itex]B_2=\{ (1,2) , (2,1) \} [/itex] is also a basis for the same vector space, and so on.

Then I realized that the point which was not clear was that if Tv is a vector in basis B then how could with respect to B' I could write P(Tv) where P is the matrix of transformation from B to B'.
When we represent a linear transform by a matrix, remember that it's only meaningful relative to some ordered basis. So, you can represent the same linear transform with respect to a different ordered basis.

Did that help in clarifying your doubt?
 
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It's possible for a vector space to have an infinite number of basis. For example, take the vector space [itex] \mathbb{R}^2[/itex]. Now [itex]B_1=\{ (0,1) , (1,0) \} [/itex] is a basis for the vector space. Notice that [itex]B_2=\{ (1,2) , (2,1) \} [/itex] is also a basis for the same vector space, and so on.



When we represent a linear transform by a matrix, remember that it's only meaningful relative to some ordered basis. So, you can represent the same linear transform with respect to a different ordered basis.

Did that help in clarifying your doubt?
The points which u mention is clear but what is unclear is the following What is unclear is that when u are doing this u are actually trying to premultiply a vector which is already in the space W. So does it mean if W ism dim space and P is n x n and m < n then when u multiply the P with Tv do u assume that the u extend the dimension of a vector in W to n by adding n -m 0 s to the end
 
  • #4
radou
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Then I realized that the point which was not clear was that if Tv is a vector in basis B then how could with respect to B' I could write P(Tv) where P is the matrix of transformation from B to B'.
x(e') = P(e)^-1 x(e), where x(e) and x(e') are representations of a vector x in basis (e), (e'), respectively, and P(e)^-1 is the transformation matrix from (e) to (e').
 

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