# Linear Transformation

## 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?

## 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:

RUber
Homework Helper
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?

Victor Feitosa
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?

RUber
Homework Helper
##[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).

Victor Feitosa
##[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..

RUber
Homework Helper
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.

Victor Feitosa
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?

RUber
Homework Helper
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.

Victor Feitosa
HallsofIvy
Homework Helper
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).

Victor Feitosa and RUber
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!