# Change of Basis wrt Linear Transformation

• May 11th 2008, 05:53 PM
joeyjoejoe
Change of Basis wrt Linear Transformation
Hi, I have one part of a two-part question related to a linear transformation.

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(a) Let a linear transformation T: R2 --> R3 be given by T[1,3]=[2,3,4] and T[-2,5]=[1,0,2]. Find the matrix of T relative to the standard basis in each space.

I actually got this part. You just have to find how T operates on [x,y] by using the fact that T is linear.

From there, it was easy to show that T[1,0] = [7/11,15/11,14/11] and that T[0,1] = [5/11,6/11,10/11].

So, the matrix of T relative to the standard bases is [ T[1,0] ; T[0,1] ].

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(b) Find a basis B of R2 and a basis C of R3 such that T relative to B,C =

[1 0]
[0 1]
[0 0]

I have a notion of what to do, but don't know how to set it up. The image under this matrix would be {[x,y,z]: z=0}, as opposed to a different planar subspace of R3 as in part (a). Do I use a change of basis procedure here? And how would I set it up?

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"He runs with it."
• May 12th 2008, 01:06 AM
Opalg
Quote:

Originally Posted by joeyjoejoe
Let a linear transformation T: R2 --> R3 be given by T[1,3]=[2,3,4] and T[-2,5]=[1,0,2].

------------

(b) Find a basis B of R2 and a basis C of R3 such that T relative to B,C =

[1 0]
[0 1]
[0 0]

I would take B to consist of the vectors [1,3] and [-2,5], and C to consist of [2,3,4], [1,0,2] and a third vector [a,b,c] chosen so as to be linearly independent of the other two vectors in C.
• May 12th 2008, 07:09 AM
joeyjoejoe
You know, that was actually one of my first thoughts about this problem. So what you are saying is:

T[x,y] = T(a[1,3] + b[-2,5]) = aT[1,3] + bT[-2,5] = a[1,3,0] + b[-2,5,0].

Then it doesn't matter that T[x,y] = [a-2b,3a+5b,0], because anything with a non-zero third coordinate would be out of the original image as found in part (a) under this basis. That makes sense. New basis. :)

I got it, thanks.