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Math Help - [SOLVED] Linear Transformations

  1. #1
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    [SOLVED] Linear Transformations

    Hi. I need help understanding this question:

    The question tells us to state why the following is not a linear transformation:

    T: R^2 \rightarrow R^2
    T(x,y) = (x^2, xy)

    Their answer is simple enough:
    There's a property: T(-v) = -T(v)
    however, T(-(1,1) ǂ -T(1,1) \Longrightarrow (by the way, is there a way to post the not equal sign in Latex. I just copy pasted from character map)
    therefore, it is not a linear transformation.

    But I am having trouble solving the question using the main two properties:
    T(a+b) = T(a) + T(b)
    T( \lambda a) = \lambda T(a)

    I started like this:

    T(x,y) = (x^2, xy)
    T(a+b) = T( \alpha_1 + \beta_1, \alpha_2 + \beta_2)=
    =( (\alpha_1 + \beta_1)^2, (\alpha_1 + \beta_1)(\alpha_2 + \beta_2))=
    = ( \alpha_1^2 + 2\alpha_1\beta_1 + \beta_1^2), ( \alpha_1\alpha_2 + \alpha_1\beta_2 + \beta_1\alpha_2 + \beta_1\beta_2)

    and from here I am having trouble proving or disproving if it equals T(a) + T(b).
    Can someone help?

    Thanks.
    Last edited by jayshizwiz; May 31st 2010 at 02:40 AM.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by jayshizwiz View Post
    Hi. I need help understanding this question:

    The question tells us to state why the following is not a linear transformation:

    T: R^2 \rightarrow R^2
    T(x,y) = (x^2, xy)

    Their answer is simple enough:
    There's a property: T(-v) = -T(v)
    however, T(-(1,1) ǂ -T(1,1) \Longrightarrow (by the way, is there a way to post the not equal sign in Latex. I couldn't find it in the tutorial. I just copy pasted from character map)
    therefore, it is not a linear transformation.

    But I am having trouble solving the question using the main two properties:
    T(a+b) = T(a) + T(b)
    T( \lambda a) = \lambda T(a)

    I started like this:

    T(x,y) = (x^2, xy)
    T(a+b) = T( \alpha_1 + \beta_1, \alpha_2 + \beta_2)=
    =( (\alpha_1 + \beta_1)^2, (\alpha_1 + \beta_1)(\alpha_2 + \beta_2))

    and from here I am having trouble proving or disproving if it equals T(a) + T(b).
    Can someone help?

    Thanks.
    The given solution used the second property, setting \lambda = -1.

    Anyway, assume the first condition holds. Then this means that T(a, b) + T(\alpha, \beta) = T(a+\alpha, b + \beta), which means that

    (a^2, ab) + (\alpha^2, \alpha\beta) = ((a+\alpha)^2, (b+\beta)^2).

    Do you understand how I got to this point? Can you see how to show that this gives you a contradiction? (Essentially, (p+q)^2 \neq p^2 + q^2.)

    Also, I would like to point out that a map can fulfill one condition but not the other.
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  3. #3
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    Thanks for such a quick response. I must've edited my post 5 times while you answered, so my original post is a bit different than what you read.

    Anyway, assume the first condition holds. Then this means that , which means that

    .

    Do you understand how I got to this point? Can you see how to show that this gives you a contradiction? (Essentially, .)

    I can see. Your technique is much better than my book's technique.

    you did:
    and got
    as in, you did both sides at once to prove they are unequal.

    The way my book answers the questions is by starting with the first step: T(a+b), doing all the work from that, and finally coming to the conclusion that it equals T(a) + T(b). For example, the way I tried to answer was like this:
    T(a+b) = T( \alpha_1 + \beta_1, \alpha_2 + \beta_2)
    =( (\alpha_1 + \beta_1)^2, (\alpha_1 + \beta_1)(\alpha_2 + \beta_2))=
    = ( \alpha_1^2 + 2\alpha_1\beta_1 + \beta_1^2), ( \alpha_1\alpha_2 + \alpha_1\beta_2 + \beta_1\alpha_2 + \beta_1\beta_2)

    and now I am having trouble proving or disproving if this equals T(a) + T(b).

    Also, I would like to point out that a map can fulfill one condition but not the other.
    It can, but it would not be a linear transformation
    Correct?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by jayshizwiz View Post
    Thanks for such a quick response. I must've edited my post 5 times while you answered, so my original post is a bit different than what you read.




    I can see. Your technique is much better than my book's technique.

    you did:
    and got
    as in, you did both sides at once to prove they are unequal.

    The way my book answers the questions is by starting with the first step: T(a+b), doing all the work from that, and finally coming to the conclusion that it equals T(a) + T(b). For example, the way I tried to answer was like this:
    T(a+b) = T(\alpha_1 + \beta_1, \alpha_2 + \beta_2)=
    =((\alpha_1 + \beta_1)^2, (\alpha_1 + \beta_1)(\alpha_2 + \beta_2))=
    = (\alpha_1^2 + 2\alpha_1\beta_1 + \beta_1^2), (\alpha_1\alpha_2 + \alpha_1\beta_2 + \beta_1\alpha_2 + \beta_1\beta_2)

    and now I am having trouble proving or disproving if this equals T(a) + T(b).


    It can, but it would not be a linear transformation
    Correct?
    Yes. My point is that what you are trying to do might have been fruitless...!

    I am a tad confused with what your problem is. Can you not just do what I did above?
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  5. #5
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    Sorry, again I edited it while you were responding. I should probably click post preview before I click post.

    The point is I didn't know how to go from

    (), () to T(a) + T(b).

    If I just would've opened up T(a) + T(b), I would've seen that there are no 2alphabeta things in it, which means it is not a linear combination.

    My book never opens up T(a) +T(b). They just do everything in T(a+b) and reorganize to show that it equals T(a) +T(b).

    It's kinda hard to explain, but the point is, that this thread is solved!

    Thanx!!
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  6. #6
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    Actually, it's almost a solved thread. Just one thing remaining...

    is there latex code for ǂ ???
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by jayshizwiz View Post
    My book never opens up T(a) +T(b). They just do everything in T(a+b) and reorganize to show that it equals T(a) +T(b).
    ...but, it doesn't equal T(a)+T(b)...
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  8. #8
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    ...but, it doesn't equal T(a)+T(b)...
    Correct, so I guess that's why the book didn't use that method, it just inserted numbers and proved that it is not true...

    So thanks for showing me a different way to answer this type of question instead of just inserting numbers.
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  9. #9
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    You answered the thread while I was in the middle of posting something else, so you may have skipped over it:

    is there latex code for the symbol ǂ ???
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by jayshizwiz View Post
    You answered the thread while I was in the middle of posting something else, so you may have skipped over it:

    is there latex code for the symbol ǂ ???
    Yeah, \neq. I find this site quite useful.
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