# Thread: Change of Basis wrt Linear Transformation

1. ## 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?

"Here's a list of appliances my father can't use."
"How did he get in trouble with a whisk?"
"He runs with it."

2. 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.

3. 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.