Formula for T with respect to Linear Transformations

Click For Summary

Homework Help Overview

The discussion revolves around a linear transformation T defined on the space of polynomials P2. The original poster is attempting to understand the transformation's formula and how to verify its properties using matrix representation.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find the matrix representation of the transformation T and verify its application on vectors. Some participants question the clarity of the transformation's definition and the notation used. Others suggest reformatting the polynomial expressions for better understanding.

Discussion Status

Participants are actively engaging with the original poster's queries, providing tips on formatting and clarifying the transformation's structure. There is an ongoing exploration of how to express the transformation correctly and verify its properties, but no consensus has been reached on the correct formulation yet.

Contextual Notes

There are mentions of formatting issues related to the forum's input system, which may be affecting the clarity of mathematical expressions. The original poster is also seeking guidance on how to prove the transformation for all polynomials in P2.

Ptricky
Messages
5
Reaction score
0

Homework Statement


Let T:P2 -> P2 be the linear operator by
T(a0 +a1x + a2x = ao + a1 (x - 1) + a2 (x-1)2


Homework Equations


part a ask to find the matrix [T]B - did, see below
part b ask to verify matrix [T]B satisfies every vector for [T]B [X]B = [T(X)]B?


The Attempt at a Solution


Matrix [T]B = 1 -1 1
0 1 2
0 0 1

What is the given formula for T? I do not see this in the book to help solve this?
 
Physics news on Phys.org
I can't tell what you're trying to say here.
Let T:P2 -> P2 be the linear operator by
T(a0 +a1x + a2x = ao + a1 (x - 1) + a2 (x-1)2


Is the left side in your formula for your transformation T(a0 + a1x + a2x2)?

Tip: Instead of typing in and , use the forum user interface to help you with this. Just below the input text box, there's a button labelled Go Advanced. After you click this button, there are a slew of buttons to help you format your work. The X2 can be used for exponents, and the X2 button can be used for subscripts. The exponents and subscripts can be anything, not just 2.
 
Welcome to PF!

Hi Ptricky! Welcome to PF! :smile:

(and yes, do try using the X2 and X2 tags just above the Reply box :wink:)
Ptricky said:
Let T:P2 -> P2 be the linear operator by
T(a0 +a1x + a2x = ao + a1 (x - 1) + a2 (x-1)2


Write each element as a vector: (a0, a1, a2) …

then what is T of (a0, a1, a2) ? :smile:
 
Mark44

Yes, that is the correct left side of equation.

Don't mean to seam like an idiot but where again am I supposed to find this "go advanced" tab? Just below the "input text box" my choices are "submit reply" and "preview post". Below that is "additional Options", I do not see "go Advanced" - sorry!

I have all the little icons right of "fonts and sizes" but none are "go advanced"?
 


Tiny-Tim,

Thanks for the welcome! My tags are up, I see the difference in options from the quick reply options!

I did 'click' the tags for super and sub. The notation shown is what I get and . Any explanation?
 
Hi Ptricky! :smile:
Ptricky said:
I did 'click' the tags for super and sub. The notation shown is what I get and . Any explanation?

erm :redface:

you have to put something inside the tag! :biggrin:
(eg P2 P2 :wink:)
 
Tiny-Tim,

figured out the formula input notation! I will retype my question.
 
Does

T(x) = ao + a1x -a1
;a2 x - 2a2 + a2

if so how do I prove this for every x = a0 + a1x + a2x2 in P2?
 
Hi Ptricky! :wink:
Ptricky said:
Does

T(x) = ao + a1x -a1
;a2 x - 2a2 + a2

No … try again!

And when you've got it right, put it in the form b0 + b1x + b2x2 :smile:
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K