# Matrix of Linear Transformation

• October 23rd 2010, 09:08 AM
dan213
Matrix of Linear Transformation
Determine the matrix representation [T]CB (C is supposed to be the subscript and B the superscript here, sorry) for the linear transformation T and ordered bases B and C.

T: P3 --> P2 given by T(p(x)) = p’(x),
B = {1, x, x^2, x^3}; C = {1, x, x^2}

I really have no idea how to proceed - the textbook is not being very helpful. I would appreciate an explanation rather than a solution, since I have many more of these problems to do. I hope someone can help!
• October 23rd 2010, 09:16 AM
TheEmptySet
Remember that to find the matrix of linear transformation
$T:\mathcal{A} \to \mathcal{B}$

Let $e_1,e_2,....,e_n$ be a basis for $\mathcal{A}$
then
$T(e_1),T(e_2),...,T(e_n)$ are the columns for the matrix representation of T when expresses in terms of the basis for $\mathcal{B}$

This is what I used in the previous problem I helped you with. I hope this gets you started.
• October 23rd 2010, 06:06 PM
dan213
So at this point I have e1 = 1, e2 = x and so on for P3. My problem is I have lost track of the geometric meaning of what I am doing, and so I can't see where I am and where I need to go in order to answer the question. If I have these column vectors how to I put them into terms of P2? I know that the kernel will annihilate at least one dimension since it is going into P2 . . .
• October 24th 2010, 09:12 AM
TheEmptySet
Quote:

Originally Posted by dan213
So at this point I have e1 = 1, e2 = x and so on for P3. My problem is I have lost track of the geometric meaning of what I am doing, and so I can't see where I am and where I need to go in order to answer the question. If I have these column vectors how to I put them into terms of P2? I know that the kernel will annihilate at least one dimension since it is going into P2 . . .

Good. So the basis of $\mathcal{A}=\{1,x,x^3,x^3\}=\{e_1,e_2,e_3,e_4\}$ and the basis of $\mathcal{B}=\{1,x,x^2\}=\{e_1,e_2,e_3\}$

Now $T(e_1)=\frac{d}{dx}(1)=0=0e_1+0e_2+0e_3$
$T(e_2)=\frac{d}{dx}(x)=0=1e_1+0e_2+0e_3$

This is the first two columns just keep going....
• October 24th 2010, 09:16 AM
dan213
Thanks, I was able to figure it out last night with your help on the previous thread. I really appreciate it.