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Help with linear transformation problem with variables

  1. May 19, 2014 #1
    Let L: R3 -> R3 be L(x)=

    [tex]
    \begin{pmatrix}
    x1+x2\\
    x1-x2\\
    3x1+2x2
    \end{pmatrix}
    [/tex]

    find a matrix A such that L(x)=Ax for all x in R2

    From what I understand I need to find the transition matrix from the elementary to L(x). However it is'nt a square matrix and it has variables instead of numbers so it confuses me. Then I should multiply that by something, which I don't know exactly what is...

    What should I do to solve this problem?
     
    Last edited: May 19, 2014
  2. jcsd
  3. May 19, 2014 #2
    Before answering, one rather urgent issue is raised: the linear transformation unaffects the condition of the input (e.g. takes a 2-D vector and gives out a 2-D vector)

    But then you define the transformation to spit out a vector that has 3-components? I must take that to mean that there is accidental error in the premise where you had rathered L(x) be from R squared to R cubed.
     
  4. May 19, 2014 #3
    Yeah my bad, sorry for the typo.
     
  5. May 19, 2014 #4
    I do not know what the term matrix means to you, but in brief and for our purposes, a matrix is simply a "linear transformation" which when acting on a vector gives out another vector! The size and shape of the matrix tells you the quality of vectors it takes in and those which it spits out (e.g. a 3x2 matrix takes in 2-D vectors and spits out 3-D vectors). The components of the matrix are to be inferred as the exact processes that alter the input vector so.

    This transformation is linear for the reason that when it acts on a sum of vectors, it may superimpose their individual results together, and when acting on a multiple of a vector, is equivalent to acting on the vector and stretching by the same constant.

    Somehow we have to exhibit some mechanism, some matrix which when acting on our 2-D vector, with components x1 and x2, gives out a 3-D vector of the specified shape.

    This is best done as follows:

    tutor1.jpg

    Note in the last step I abused the following fact (one that is to be well understood)- a column vector which when applied by a matrix, reproduces another column vector that takes a linear combination of the columns of the matrix.
     
    Last edited: May 19, 2014
  6. May 19, 2014 #5
    While the above technique is the most promising, as it makes reference to no other special theorem by itself, an alternative popular approach is as follows:

    ask yourself what does this transformation do on the vector (1,0)?

    what does it do to the other basis vector (0,1)?

    These form the columns of the transformation matrix! The reason why has intimately to do with the priori the transformation is linear!
    tutor2.jpg
     
  7. May 19, 2014 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    (The first post says "R3 to R3" but sudhirkings quote says "R2[/tex]". I will assume R3 is intended.)

    So L maps vector [tex]<x_1, x_2, x_3>[/tex] to [tex]<x_1+ x_2, x_1- x_2, 3x_1+ 2x_2>[/tex] and you want to write it as a 3 by 3 matrix? In order to do this you need to specify the bases so I will assume the basis is the "standard" basis {<1, 0, 0>, <0, 1, 0>, <0, 0, 1>}.

    Apply L to each of the basis vectors in order. Write the result as a linear combination of the basis vectors. The coefficients form the columns of the matrix representation.

    For example, here, L(<1, 0, 0>)= <1+ 0, 1- 0, 3(1)+ 2(0)>= <1, 1, 3>= 1<1, 0, 0>+ 1<0, 1, 0>+ 3<0, 0, 1> so the first column is [itex]\begin{bmatrix}1 \\ 1 \\ 3\end{bmatrix}[/itex].
     
  8. May 20, 2014 #7


    I actually believe the OP made a slight typo again! I believe he wishes the transformation to be from 2-D to 3-D.

    Reason I infer this is from the following:

    * the initial post posited that the transformation was from 2-D to 2-D (I think in my first post you can see it in the quote).

    *The OP goes on to ask what is "L(x) where x is in R^2" or something to the same effect.
     
    Last edited: May 20, 2014
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