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Linear Algebra: Is this a linear transformation?

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Is T a linear transformation?

    T: M22 --> M22 defined as:

    T [ a b ] = [ 1 (a-d) ]
    , [ c d ] ,,, [ (b-c) 1 ]


    2. Relevant equations

    none.

    3. The attempt at a solution

    I need to show that it is closed under addition and scalar multiplication. This is as far as I could get it to:

    T [ a b ] + [a' b' ] = T [ (a+a') (b+b') ] = [ 1 (a+a')-(d+d') ]
    , [ c d ] ,,, [c' d' ],,,, [ (c+c') (d+d') ] ,,,, [ (b+b')-(c+c') 1 ]

    and now I'm kind of stuck. I don't see how that last one could be turned back into its original transformation.

    (please ignore the commas because I was trying to make the matrices readable)
     
    Last edited: Feb 26, 2009
  2. jcsd
  3. Feb 26, 2009 #2

    J$C

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    Well the question asks if it is a linear transformation. Try to find a counter-example and you will see why you can't prove it. Notice the transformation always turns (1,1) and (2,2) into 1.
     
  4. Feb 26, 2009 #3

    lanedance

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    isn't T a tranformation on a matrix?

    so i think you have taken T.(A+A') where a is the matrix

    to show closure under addition & mult you need show.. or not

    T.k(A+A') = k(T.A) + k(T.A')
     
  5. Feb 27, 2009 #4

    J$C

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    The key is that for matrices A and A', T(A+A') will have ones in the upper left and lower right entries by the nature of the transformation. T(A), and T(A') will also have that. So look at these entries and compare them on T(A+A') and T(A)+T(A').
     
  6. Feb 27, 2009 #5

    lanedance

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    might be alot easier to try multiplication first

    ie see if T.(kA) = k(T.A) for some scalar k
     
  7. Feb 27, 2009 #6

    HallsofIvy

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    Even more specifically, compare
    [tex]T\left[\begin{array}{cc} 2 & 0 \\ 0 & 2\end{array}\right][/tex]
    and
    [tex]2T\left[\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right][/tex]
     
  8. Feb 27, 2009 #7
    Wow didn't expect so many responses.

    I can see that it won't work for scalar multiplication because of those ones in (1,1) and (2,2)

    Those are all referring to 1, so then T(A+A') and T(A)+T(A') are not equal in the (1,1) and the (2,2) positions. Since they're not equal, that must mean that they're not closed right?
     
  9. Feb 27, 2009 #8
    Yes.

    Just concerning the terminology: If you want to show that some subset of a vector space is a vector subspace you have to show it is closed under addition and scalar multiplication. On the other hand, a map is linear if it commutes with addition and scalar multiplication. A function which commutes with addition is also called "additive".
     
  10. Mar 2, 2009 #9
    Awesome thanks everyone.

    Just one more question, someone else also told me that if the zero vector is not included in the vector space during a transformation, then it is not a linear transformation. That would be a really quick way to find out if it's linear, but I just don't see how or why that is. I thought anything could be "moved" to make it go through the zero vector?
     
  11. Mar 2, 2009 #10

    HallsofIvy

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    I don't know what you mean by "zero vector is not included in the vector space during a transformation". It is certainly true that T(0)= 0 for every linear transformation so the 0 vector must be in the range (image) of any linear transformation. Was that what you mean?
     
  12. Mar 2, 2009 #11

    J$C

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    What they probably meant is since T(0)=0, if you can quickly see that 0 is not in the range then certainly it's not going to be a linear transformation because T(0)=something else. If T(0)=a and a,b are nonzero then b*T(0)=b*a, but if T is linear then b*T(0)=T(b*0)=T(0)=a [tex]\neq[/tex]a*b.

    Or a more compact way to think about it, T(0)=T(a*0)=0*T(a)=0.

    So, considering it a linear transformation would fall apart unless T(0)=0, or even more generally it won't work it 0 is not in the range, because then clearly T(0) cannot be equal to 0.
     
    Last edited: Mar 2, 2009
  13. Mar 5, 2009 #12
    That's exactly what I wanted to know! :)

    and this helps clear it up! I was trying to look for little things to speed through some questions.
    happy happy mood. :smile:
    Thank you!!!
     
    Last edited: Mar 5, 2009
  14. May 9, 2009 #13
    Hi ! My name is Jennifer. I'm new to Physics Forum. I was using google to help me solve an algebra question and I came across this site. I'm not sure how to post a question so I'm putting it here....so sorry. But this is the question:-

    Let L : Rn --> Rm and M : Rm --> Rp be linear mappings.
    a) Prore that rank( M o L) <= rank(M).
    b) Prove that rank( M o L) <= rank(L).

    Thank you~
    Jennifer
     
  15. May 11, 2009 #14

    lanedance

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    Hi Jennifer1990 - welcome to PF

    you will probably have more luck getting a response by posting the the "Homework & Coursework Questions > Calculus & Beyond" as a fresh topic... people are also usually more helpful if you provide some attempt or idea to start at the problem - any ideas?

    as a starting point, one defintion of rank of a linear map is the dimension of the image of the map ...
     
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