# Not clear about the change of basis in new space

1. Apr 6, 2007

### Sumanta

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.

2. Apr 6, 2007

### siddharth

It's possible for a vector space to have an infinite number of basis. For example, take the vector space $\mathbb{R}^2$. Now $B_1=\{ (0,1) , (1,0) \}$ is a basis for the vector space. Notice that $B_2=\{ (1,2) , (2,1) \}$ 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?

Last edited: Apr 6, 2007
3. Apr 6, 2007

### Sumanta

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. Apr 6, 2007