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Math Help - image of T with by some basis

  1. #1
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    image of T with by some basis

    \begin{pmatrix}<br />
4 &2 &0| &a1 \\ <br />
2 & 3 &2| &a2 \\ <br />
0 &2 &2| &a3 <br />
\end{pmatrix}
    i need to solve it find (x,y,z)
    replace x y and z with expression of a1 a2 a3 and get the span of the image in some basis

    but after doing somw row reduction i get a line of zeros equals some exprression

    and here y is a free variable so i cant get expressiom of a1 a2 a3 of it

    what to do?

    i need only a1 a2 a3 reprentation

    having free variables messes it up
    Last edited by transgalactic; July 4th 2011 at 03:23 PM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: image of T with by some basis

    Why don't you transcribe the entire question?
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  3. #3
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    Re: image of T with by some basis

    oohh sorry i wrote the wrong question.

    here is the original question.

    there is A 2x2 which is
    (1 -1)
    (-1 1)

    T:M 2x2 (R)-> M 2x2(R)
    we definf the transformation
    T=AM
    find ImT ??

    i know there is another way.
    but i want to solve it in this perticular way.

    i got the formula for the transformation
     <br />
T=\begin{pmatrix}<br />
x &y \\ <br />
z &t <br />
\end{pmatrix}=\begin{pmatrix}<br />
x-z &y-t \\ <br />
-x+z &-y+t <br />
\end{pmatrix}<br />

    i founded the representative matrice by the standart basis
    and i made it equal to some (a1,a2,a3,a4) vector

     <br />
\begin{pmatrix}<br />
1 &0 &-1 &0 &a1 \\ <br />
0 &1 &0 &-1 &a2 \\ <br />
-1 &0 &1 &0 &a3 \\ <br />
0 &-1 &0 &1 & a4<br />
\end{pmatrix}<br /> <br />
    affter row reduction

     <br />
\begin{pmatrix}<br />
1 &0 &-1 &0 &a1 \\ <br />
0 &1 &0 &-1 &a2 \\ <br />
0 &0 &0 &0 &a3+a1 \\ <br />
0 &0 &0 &0 & a4+a2<br />
\end{pmatrix}<br />

    replace x y z t with expression of a1 a2 a3 a4 and get the span of the image in some basis

    but after doing somw row reduction i get a line of zeros equals some exprression

    and here z t is are free variable so i cant get expressiom of a1 a2 a3 a4 of it

    what to do?
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  4. #4
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    Re: image of T with by some basis

    ok, if i understand you correctly, T is defined on M_{2x2}(\mathbb{R}) by: T(U) = AU, where U is any 2x2 real matrix, and A =

    [1 -1]
    [-1 1]. now T is clearly not 1-1 since:

    [a a]
    [a a] is in ker(T), for any a in R (even no-zero numbers). if we choose the standard basis \{E_{ij}\} for M_{2x2}(\mathbb{R}),

    we get the 4x4 matrix you indicate, which has rank 2. this means, among other things, that we need to only specify

    2 linearly independent elements of M_{2x2}(\mathbb{R}) in im(T) to obtain a basis.

    now in terms of x,y,z,t this becomes:

    [1 -1][x y]....[a1 a2]
    [-1 1][z t] = [a3 a4]. remember, x, y, z and t can be ANY real numbers, we're not "solving" for them.

    what we wind up with is:

    [x-z y-t]....[a1 a2]
    [z-x t-y] = [a3 a4], so a3 = -a1, a4 = -a2. so any element of im(T) is of the form:

    [ a1 a2 ]
    [-a1 -a2]. so the only question is, is every matrix of this form in im(T)? well suppose we have a matrix of that form.

    we can set x = a1, y = a2, z = t = 0 so the matrix:

    [a1 a2]
    [ 0. 0 ] is clearly a pre-image.

    to answer your question in the terms you asked it, since z,t are "free" pick some convenient values for them (0 works well, here, but choosing

    some other values would just yield a different basis for im(T) ). hopefully you see we can write

    im(T) as span( \{E_{11}-E_{21},E_{12}-E_{22}\})
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  5. #5
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    Re: image of T with by some basis

    ok bu if i wiil take the other method(the formal solution
    i get four 2x2 matrices
    so there is a single expression for x y z t

    how to get the same result in my method
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  6. #6
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    Re: image of T with by some basis

    in the screenshot you attached, you have pretty much just what i wrote (replace "x" with "a", "y" with "b", "z" with "c", and "t" with "d").

    the two matrices listed as the spanning set, are the same two matrices in my spanning set, just written a different way. for example E11 =

    [1 0]
    [0 0], and E21 =

    [0 0]
    [1 0], so E11-E21 is the first matrix you have shown as a basis element of Im(T).

    if you regard T as a 4x4 matrix R^4-->R^4, then your row-reduction shows the first two columns of T form a basis for Im(T):

    Im(T) = span({(1,0,-1,0), (0,1,0,-1)}), because these are the pivot columns for the matrix of T. converting these elements of R^4

    back into 2x2 matrices gives the same answer as in the book.
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