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

AI Thread Summary
The discussion focuses on solving a linear transformation problem involving the mapping of vectors in R2. The user initially struggles with expressing the vector (-1,5) as a linear combination of given vectors and attempts to apply the linear transformation rules. After deriving coefficients for the linear combination, they correctly apply the transformation to find L(-1,5) but encounter difficulties when trying to generalize the solution for L(a1,a2). The conversation reveals confusion about the correct approach to solving for a and b, leading to incorrect transformation results. The user seeks clarification on their methodology and the correct steps to achieve the desired outcome.
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?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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