Linear Transformation Part 2: Finding the Image of a Linear Transformation

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Homework Help Overview

The discussion revolves around a linear transformation from R² to R², specifically focusing on finding the images of given vectors under this transformation. The original poster presents known transformations of specific vectors and seeks to determine the transformation of other vectors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore expressing vectors as linear combinations of known vectors and applying the properties of linear transformations. Questions arise regarding the correct setup for finding coefficients and the application of linearity.

Discussion Status

Participants are actively engaging with the problem, attempting to derive expressions for the transformation of new vectors based on previous results. Some guidance has been offered regarding the need to solve for coefficients before applying the transformation, but no consensus on the correct approach has been reached.

Contextual Notes

There are indications of confusion regarding the application of linear combinations and the derivation of coefficients, as well as discrepancies in matrix manipulation. Participants are also questioning the assumptions made in their calculations.

aznkid310
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Homework Statement



Let R2 => R2 be a linear transformation for which we know that:

L(1,1) = (1,-2)
L(-1,1) = (2,3)

What is: L(-1,5) and L(a1,a2)?


Homework Equations



I don't know where to start. I tried writing (-1,5) as a linear combo of (1,1) and
(-1,1), but that got me nowhere. Am i suppose to find a basis? How do i do that?

The Attempt at a Solution



(-1,5) = a(1,1) + b(-1,1)

Transforming it into a 2 x 3 matrix and row reducing it, i get a = 2, b = 3.
If i am on the right track, what do i do next?
 
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If (-1,5)=2*(1,1)+3*(-1,1), then L(-1,5)=2*L(1,1)+3*L(-1,1), right? That's what 'linear' is all about. Now do the same thing for (a1,a2).
 
Ok i think i got part a!

For L(a1,a2), i stuck at: L(a1,a2) = 2(L(a1,a2)) + 3(L(a1,a2))

I was looking at some of examples, and i had no idea where they get the (a+b)/2 and (a-b)/2 terms from.
 
Aren't you going to solve (a1,a2)=a*(1,1)+b*(-1,1) first for a and b, so you can do EXACTLY the same thing as you did for part a? Except this time a=2 and b=3 are changed. I think that's the pedagogical point.
 
Last edited:
so this is what i did:

[ 1 -1 a1
1 1 a2 ]

[ 1 -1 a1
0 2 a2 ]

which means b = a2/2, a = a1 + a2/2

Then L(a1,a2) = (a1 + a2/2)( L(1,1)) + (a2/2)(L(-1,2))

= (a1 + a2/2)(1,-2) + (a2/2)(2,3)

= (a1 + a2/2, -2a1 - a2) + (a2, (3a2)/2)

= (a1 + (3a2)/2), (-2a1 + a2/2)

But, this is not the right answer. What did i do wrong?
 
aznkid310 said:
so this is what i did:

[ 1 -1 a1
1 1 a2 ]

[ 1 -1 a1
0 2 a2 ]
This is wrong. You subtracted the first row from the second so this should be
[ 1 -1 a1
0 2 a2-a1]
Of course, what you are really saying is that a- b= a1 and a+ b= a2. That should be easy to solve without matrix methods.
which means b = a2/2, a = a1 + a2/2

Then L(a1,a2) = (a1 + a2/2)( L(1,1)) + (a2/2)(L(-1,2))

= (a1 + a2/2)(1,-2) + (a2/2)(2,3)

= (a1 + a2/2, -2a1 - a2) + (a2, (3a2)/2)

= (a1 + (3a2)/2), (-2a1 + a2/2)

But, this is not the right answer. What did i do wrong?
 

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