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Homework Help: Ok i got to the end of this whole Im theory but

  1. Feb 9, 2008 #1
    ok i got to the end of this whole Im theory but....

    i have two cases that i need guidense

    in the link i showed a basic case and how we solve it using this method


    than i presented the first case that i cant solve
    i which i tried to copy a case that we have delt with before
    (with the b2=0)

    than i presented the second case that i cant solve like our (b2=0) example

    basicly may method say to make A*x=b

    then to find the "b"s value for which we get a solution for the matrix

    in the end we copy the "b" column with the conclution the gives us an answer for the matrix.

    in both cases i get to part of the conclution but i cant go further
    to wright down the "b" column with this conclutions and to
    exract the vectors of the basis


    (by the way)
    i am using this method because i was tought that it works for every basis
    unlike the tangugation mathod that works only for standart basis
    Last edited: Feb 9, 2008
  2. jcsd
  3. Feb 9, 2008 #2


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    Science Advisor

    How many different problems do you have here?

    The first appears to be "Find the kernel of linear transformation A where A is the matrix
    [tex]\left[\begin{array}{cccc} 1 & 0 & 0 & -3 \\ 2 & 0 & 4 & -5 \end{array}\right][/tex].
    from R4 to R2. It is easy to see that, multiplied by [itex]<x_1, x_2, x_3, x_4>[/itex] gives <x_1- 3x_4, 2x_1+ 4x_3- 5x_4>[/itex] it is equally easy to see that, for arbitrary [itex]x_1, x_2, x_3, x_4[/itex] those can give any numbers. The image of A is all of R2 and so < 1, 0> and <0, 1> are basis vectors. The rest of what you have there, I simply don't understand. And I have no idea what "tangugation" means! Translate please.
  4. Feb 9, 2008 #3
    all of my questions are about finding the Im

    i have there two cases that i dont know how to find their Im

    i also tried to solve them using your method
    but i have troubles with it


    and tangugation i think its the operation of transforming column into rows

    i dont have a problem with the first metrix
    i just showed an example of how this method is working
    the problem is with the second metrix and the last metrix
    in the middle i gave another example of how to find the Im
    using this method
    Last edited: Feb 9, 2008
  5. Feb 9, 2008 #4


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    Your first operator changes the general (x, y, z, t) into (0, -y+ z, 0, 0). Isn't it obvious, that we can pick y and z to make -y+ z anything? (More technically, we can always solve -y+ z= a for any a- and it has an infinite number of solutions.) The image is simply the one dimensional subspace (0, a, 0, 0). You were almost there in writing y(0, -1, 0, 0)+ z(0, 1, 0, 0) since that is equal to (-y+ z)(0, 1, 0, 0). The image is the subspace having basis (0, 1, 0, 0) and, of course, is one-dimensional.

    But you say the answer in your book is (0, 1, -1, 0). I don't know how they got that. The "augmented" matrix you show
    [tex]\left(\begin{array} {ccccc}0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & b_2 \\ 0 & 0 & 0 & 0 & b_3+ b_2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)[/tex]
    Looks like it is the result of row reducing another augmented matrix: it looks like it started as
    [tex]\left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & b_2 \\ 0 & 1 & -1 & 0 & b_3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)[/tex]
    In other words, that the original transformation "mapped" (x, y, z, t) to (0, -y+ z, y- z, 0). If that is the case,since y- z= -(-y+ z) and, again, for y and z arbitrary, y- z can be anything, that is of the form (0, a, -a, 0)= a(0, 1, -1, 0) and, in this case, the image is the subspace having basis {(0, 1, -1, 0)}.

    So which is, in fact, the correct transformation? (x, y, z, t)-> (0, -y+ z, 0, 0) which is what you give initially, or (x, y, z, t)-> (0, -y+ z, y- z, 0) which is what your book seems to be working with.

    For the second problem where T(x, y, z)= (2x- y- z, -y+ z, 0), since 2x-y-z and y- z are arbitrary (there exist at least one solution to 2x- y- z= a and y-z= b for all a and b), any vector in the image if of the form (a, b, 0)= a(1, 0, 0)+ b(0, 1, 0). The image has basis {(1, 0, 0), (0, 1, 0)}.

    Swapping rows and columns, in English, at least, is "transposition". What by, the way, is your native language? Your English is excellent.
    Last edited by a moderator: Feb 9, 2008
  6. Feb 9, 2008 #5
    ok i understood your method
    the only problem that remains for me is
    "when to start this proccess???"

    of making A*(x,y,z)

    in any general problem i get a metrix like

    (a b c)
    (d e f)
    (s t y)

    from substituting A(e1) A(e2) A(e3) in to the general formula

    than i try to transform it so it will contain as much zeros as possible

    (a b c)
    (0 e f)
    (0 0 y)

    it can take several operations of row reduction for me to get to this resolt
    i can go further and make this metrix to look like this

    (a b 0)
    (0 e 0)
    (0 0 y)

    i got the impression that we can get many Im resolts
    because our matrix can change many time.
    for example when you worked with
    0 0 0 0
    0-1 1 0
    0 0 0 0

    you got an answer that differs the answer in my book
    so you took the previos form of this matrix (one step before)
    and then you made A*(x,y,z) and then you got the right answer

    what is the guide line for me to know
    for stopping row reduction procces and starting with A*(x,y,z) proccess

    Last edited: Feb 9, 2008
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