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Math Help - Find the basis of a linear transformation

  1. #1
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    Find the basis of a linear transformation

    Hi,

    I've got an exam tomorrow and I'm trying to attempt past papers.
    This question is really confusing me.


    Find the basis of a linear transformation-basis.jpg




    I can do part b), c)i, c)ii, c)iii but cannot complete the rest.
    How can I find the basis of the above linear transformation?

    Any help appreciated
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  2. #2
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    Zagreb
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    Well i will try to help... ( my english is bad so if you dont understand some words just tell and i will try to explain better )
    In a) part you need to give a basis for V.
    So the "best" basis for any space is canon basis
    ( example.
    for space R^2 the best basis is:
    { ( 1, 0 ) , ( 0 , 1 ) }
    )

    So in this case the best basis would be:

    1 0 0
    0 0 0 = A1
    0 0 0

    0 1 0
    0 0 0 = A2
    0 0 0

    0 0 1
    0 0 0 = A3
    0 0 0

    0 0 0
    0 1 0 = A4
    0 0 0

    0 0 0
    0 0 1 = A5
    0 0 0

    0 0 0
    0 0 0 = A6
    0 0 1


    to show that this set ( lets name it S ) of matrices is really basis for space of upper triangular matrices you need to show this:

    if you have any upper triangular matrix:

    x1 x2 x3
    0 x4 x5 = X
    0 0 x6

    you can denote it like:

    X = x1*A1 + x2*A2 + x3*A3 + x4*A4 + x5*A5 + x6*A6

    so this is the proof that you can get any upper triangular matrix with linear combination of matrix from set S.

    now you need to show that: ( y1, ... y6 ----> some variables )

    0 0 0
    0 0 0 = y1*A1 + y2*A2 + ... + y6*A6
    0 0 0
    is only on trivial way...
    that is easy ( you need to show that this is true only if y1 = y2 = ... y6=0)
    and that is the proof that S is basis for space of upper triangular matrices



    c) iv

    dont really know what you need to do ( dont understand ), but it cant be hard and if that is what i think it is than you will know what to do ( if you really know to solve problems c)i, ii, iii )
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  3. #3
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    How did you compute ci, ii, iii? I have a similar question and struggling with the basics...
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  4. #4
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    Quote Originally Posted by math_help View Post
    How did you compute ci, ii, iii? I have a similar question and struggling with the basics...
    I agree that ci, cii, and ciii are much harder thatn a and b!

    You are given the matrix M= \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix} and your linear transformation is defined by T(A)= MA.

    Any "vector" in V is a matrix of the form \begin{bmatrix}a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix} which is
    \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\ 0 & d & e \\ 0 & 0 & 2f\end{bmatrix}

    ci) the range is the set of all possible results of that. Obviously (I hope it is obvious), that is the set of matrices of the form \begin{bmatrix}0 & 0 & 0 \\ 0 & a & b \\ 0 & 0 &  c\end{bmatrix} for some numbers a, b, c.

    cii) the kernel is the set of all matrices A such that MA= 0 or \begin{bmatrix}0 & 0 & 0 \\ 0 & d & e \\ 0 & 0 & 2f\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.
    Setting comparable components of those equal, we must have d= 0, e= 0, and 2f= 0. a, b, and c can be anything so the kernel consists of matrices of the form \begin{bmatrix}a & b & b \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}.

    ciii) The rank of T is the dimension of its image. That should be obvious: any of the three numbers a, b, c in \begin{bmatrix}0 & 0 & 0 \\ 0 & a & b \\ 0 & 0 &  c\end{bmatrix} can be chosen arbitrarily so the dimension of the image, and the rank of T is 3. (The kernel, in cii, also has dimension 3 so the "nullity" of T is 3, confirming the formula "rank(T)+ nullity(T)= dimension(V)".)

    civ) Standard way of forming the matrix corresponding to a linear transformation, in a given basis: Apply the linear transformation to each of the basis vectors in turn. Write the result as a linear combination of the basis vectors. The coefficients form a column of the matrix.

    Here: Apply T to each of the 6 basis vectors in (a). For example, the first basis "vector" josipive gave you was A_1= \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}. So T(A_1)= \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}. (Here, A_1 is in the kernel of T) That can be written 0A_1+ 0A_2+ 0A_3+ 0A_4+ 0A_5+ 0A_6 so the first column in the matrix representation of T is a column of 6 zeros.
    Last edited by HallsofIvy; May 13th 2009 at 07:28 AM.
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  5. #5
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    I'd like to say thanks for all the help.
    I was struggling with the concept of finding a basis for a matrix rather than a vector, but I now understand what to do.

    thanks again!
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  6. #6
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    So it has dimension 6? Surely looking at the question you're meant to be able to deduce this first...
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