How Do You Compute the Matrix of a Linear Map T Defined by T(P(x)) = xP(x)?

  • Thread starter Thread starter jiles-smith
  • Start date Start date
  • Tags Tags
    Linear Matrix
Click For Summary

Homework Help Overview

The problem involves computing the matrix of a linear map T defined by T(P(x)) = xP(x), where T maps polynomials from R[x]2 to R[x]3. The task includes determining the matrix representation with respect to specified bases and finding the kernel and image of T.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss computing the action of T on basis vectors and expressing results in terms of the target basis. There are questions about the correct representation of bases in vector form and how to transition between the two spaces.

Discussion Status

Some participants have proposed methods to compute the matrix and have begun to derive results based on their interpretations of the linear map. There is ongoing clarification regarding the dimensionality of the spaces involved and the correct notation for the bases.

Contextual Notes

There are mentions of confusion regarding the dimensions of the spaces and the appropriate basis representations, indicating potential misunderstandings about the problem setup.

jiles-smith
Messages
6
Reaction score
0

Homework Statement



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.


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.
 
Physics news on Phys.org
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:
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?
 
jiles-smith said:
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?
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.)
 
jiles-smith said:
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?

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

Similar threads

  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K