Linear Transformation

  • #1

Homework Statement



Being T: ℝ2 → ℝ2 the linear operator which matrix in relation to basis B = {(-1, 1), (0, 1)} IS:
[T]b =
\begin{bmatrix}
1 & 0\\
-3 & 1
\end{bmatrix}

True or False: T(x,y) = (x, 3x+y) for all x,y∈ℝ?

Homework Equations




The Attempt at a Solution

3
[/B]
So first I convert (x,y) from canonical to basis B and found (-1, 2).
Next, I calculate [T]b. (-1,2). Found that the result is (-x, 3x +2y).
Now, from my perspective i have to convert it to canonical basis again.
But how am I suposed to do it?
 
Last edited:

Answers and Replies

  • #2
RUber
Homework Helper
1,687
344
I am a little unclear of what the question is asking.
## [T]_b = \begin{bmatrix} 1&0\\-3&1 \end{bmatrix} ##
And in stacked matrix form:
## B = \begin{bmatrix} -1&0\\1&1 \end{bmatrix} ##
So, you are looking for a matrix A such that AB = [T]_b. Is that the correct understanding of the question?
 
  • Like
Likes Victor Feitosa
  • #3
Sorry, I just want to know if the following statement is true: True or False: T(x,y) = (x, 3x+y) for all x,y∈ℝ?
 
  • #4
RUber
Homework Helper
1,687
344
##[T]_b## refers to the matrix operation conducted relative to the basis set B.
So if you start with (x,y) in the standard basis, what is (x,y) in basis B? ( y-x, y).
Then apply T_b to that and you will get T(x,y).
 
  • Like
Likes Victor Feitosa
  • #5
##[T]_b## refers to the matrix operation conducted relative to the basis set B.
So if you start with (x,y) in the standard basis, what is (x,y) in basis B? ( y-x, y).
Then apply T_b to that and you will get T(x,y).

I did what you said, but I get that T(x,y) = (-x+y, 3x-2y)
How I convert it to standard basis again?
I tried to multiply T(x,y).B , but my result is (4x -3y, 3x-2y)
I don't know what I am doing wrong..
 
  • #6
RUber
Homework Helper
1,687
344
I did what you said, but I get that T(x,y) = (-x+y, 3x-2y)
This looks right.
How I convert it to standard basis again?
You don't -- x and y are representative of the standard basis.
 
  • Like
Likes Victor Feitosa
  • #7
But the answer says that the statement T(x,y) = (x, 3x+y) for all x,y∈ℝ? is True.

Also, [T]b.(x,y)b return the answer in Basis B, not?
 
  • #8
RUber
Homework Helper
1,687
344
Hmmm. Okay. So perhaps we did the operation backward to get from (x,y) in the standard basis to (a,b) in the basis B.
To point to ( x,y )_{x,y} you need a = -x, and b = x+y, or (-x, x+y)_B.
Hit that with T_b.
You get ( -x, 3x+x + y ) = (-x, 4x+y) in basis B.
Putting that back into the canonical basis, gives ( x , -x+4x+y).
Now I can see that the answer should be true.

The problem was that we were using the definition for basis vectors of B, as the operation to move from (x,y)_{x,y} to (a,b)_B, where really it is B * (a,b)_B = (x,y).
This means that:
(a,b)_B = B^{-1} (x,y).
T_b (a,b)_B = T_b B^{-1} (x,y), which as you said was in basis B.
So B T_b (a,b)_B =B T_b B^{-1} (x,y) will return the answer in the standard basis.

Sorry that took so long.
 
  • Like
Likes Victor Feitosa
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,847
967
Since you are told what T is "in relation to basis B = {(-1, 1), (0, 1)}", write (x, y) in that basis:
(x, y)= a(-1, 1)+ b(0, 1)= (-a, a+ b) so we have -a= x and a+ b= y. a= -x so a+ b= -x+ b= y, b= x+ y.

(x, y)= -x(-1, 1)+ (x+ y)(0, 1). T(x, y)= -xT(-1, 1)+ (x+ y)T(0, 1).
 
  • Like
Likes Victor Feitosa and RUber
  • #10
Thank yu both, guys!!!!!!!
It took long to understand but thanks to you all I managed to grasp it. And I think it'll stick now!
 

Related Threads on Linear Transformation

  • Last Post
Replies
1
Views
836
  • Last Post
Replies
5
Views
695
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
J
  • Last Post
Replies
2
Views
676
  • Last Post
Replies
3
Views
786
  • Last Post
Replies
6
Views
853
  • Last Post
Replies
11
Views
1K
Top