# Linear maps: finding matrix

1. Nov 3, 2009

### jiles-smith

1. The problem statement, all variables and given/known data

Let T:R[x]2->R[x]3 be defined by T(P(x))=xP(x). Compute the matrix of T with respect to bases {1,x,x^2} and {1,x,x^2,x^3}. Find the kernel and image of T.

3. The attempt at a solution
I genuinely have no idea where to start on this, any pointers you can give me would be greatly appreciated.

2. Nov 3, 2009

### lanedance

i would start be computing the action of T on each basis vector of R2 & write in terms of basis of R3

then use that to make a matrix

notice the polynomials are considered as vectors, so for example ax2 +b x + c in the basis of R2 could just be written (c,b,a)

Last edited: Nov 4, 2009
3. Nov 4, 2009

### jiles-smith

so would the basis in R^2 in vector form be: {1,0,0},{0,1,0},{0,0,1}? If so, how would you get this in terms of the R^3 basis?

4. Nov 4, 2009

### HallsofIvy

What you give, since they are in R3, form a basis for R3, not R2. A basis for R2 is {(1, 0), (0,1)}. (Don't use "{" and "}" for individual vectors. Those are set delimiters.)

5. Nov 4, 2009

### lanedance

as halls points out its not R^2, I'm not too sure what the proper notation is, but as there are 3 independent basis vectors, it much more like R^3

so for the space, like R^3 with the basis {1,x,x2}, i think you're correct that you identify
(1,0,0) with 1
(0,1,0) with x & so on

simarlarly for the space where the image resides, with the basis {1,x,x2,x^3}, I would identify
(1,0,0,0) with 1
(0,1,0,0) with x & so on

is what you wrote exactly how the question is written?

Last edited: Nov 4, 2009
6. Nov 4, 2009

### jiles-smith

okay, so i have used the basis for R^3 and got a diagonal matrix with the elements x. Given the equation T(P(x))=x P(x) it looks like it could work. Is this correct? cheers

7. Nov 4, 2009

### lanedance

what matrix do you get, the way I'm thinking there shoudn't be any x's in the matrix?

if its from a 3 dimensional space to a 4 dimensional space, i think it should be a 4x3 matrix

Also i think it should have constant entries... consider what multiplying by x does, it shifts you from one basis vector to another... (similar to 90degree rotation in normal R^3)

Last edited: Nov 4, 2009
8. Nov 4, 2009

### jiles-smith

well i have just changed direction a bit but here is what i have so:
T(1,0,0)=(0,1,0,0); T(0,1,0)=(0,0,1,0); T(0,0,1)=(0,0,0,1)
So the matrix would be:
((0,0,0),(1,0,0),(0,1,0),(0,0,1))

(Each sub-bracket is a row)

9. Nov 4, 2009

### lanedance

sounds reasonable to me & all lines up with the initial definition