Linear Transformation: Converting Between Canonical and Basis Representations

[Victor Feitosa]

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?

 

[RUber]

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]

$[T]_b$ 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]

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]

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.

 

[HallsofIvy]

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]

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!