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Math Help - Getting the image: involves LT composition and change of basis matrix.

  1. #1
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    Getting the image: involves LT composition and change of basis matrix.

    Hi!

    i have a doubt here:

    Excercise:
    Given:
    B_{1}=\{v_{1},v_{2},v_{3}\}
    B_{2}=\{-v_{2}-v_{3},v_{1}+v_{2}+v_{3},v_{1}-v_{2}\}
    Both are basis from the V linear space.
    and
    f:V \rightarrow V \ / \ M(f)_{B_{1}B_{2}}= \left(\begin{array}{ccc}1&-2&1\\2&1&-2\\-1&-3&3\end{array}\right)

    -->Find Image(fof)



    So, my plan was:

    Get the change of basis matrix, this is easy because of the way the basis are expresed
     C_{B_{2}B_{1}} = \left(\begin{array}{ccc}0&1&1\\-1&1&-1\\-1&1&0\end{array}\right)

    Now i know that C_{B_{2}B_{1}}.M(f)_{B_{1}B_{2}}=M(f)_{B_{1}B_{1}}

    \left(\begin{array}{ccc}0&1&1\\-1&1&-1\\-1&1&0\end{array}\right) \left(\begin{array}{ccc}1&-2&1\\2&1&-2\\-1&-3&3\end{array}\right) = <br />
\left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right)

    Now i just do
    M(f)_{B_{1}B_{1}}M(f)_{B_{1}B_{1}}
    \Rightarrow \left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right) \left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right) = \left(\begin{array}{ccc}-2&-11&10\\8&14&-16\\4&7&-8\end{array}\right)

    And those columns are the image of  (fof) but expresed in B_{1}, so, image well expresed is
     \ Image = < -2v_{1} + 8 v_{2} + 4v_{3} ;  -11v_{1} + 14v_{2} + 7v_{3} ; 10v_{1} -16v_{2} - 8v_{3} > butīs LD, so
     \ Image = \{ -2v_{1} + 8 v_{2} + 4v_{3} ;  -11v_{1} + 14v_{2} + 7v_{3} \}
    -----------------------------------------

    So, the problem is that in the book i found a proposition that says that

    given linear functions  f:V \rightarrow W \ and  g:W \rightarrow U

    \Rightarrow \ M(gof)_{B_{1}B_{3}} = M(g)_{B_{2}B_{3}}.M(f)_{B_{1}B_{2}}

    -----------------------------------------------------

    So, that, what i have done is:  M(f)_{B_{1}B_{2}}.M(f)_{B_{1}B_{2}} = M(fof)_{B_{1}B_{2}} = \left(\begin{array}{ccc}-4&-7&8\\6&3&-6\\-10&-10&14\end{array}\right)

    and i know now that the image of the LT are the colums of the matrix, but those are expresed in  B_{2} so, the image is going to be formed by:

     \{ -4(-v_{2}-v_{3}) + 6(v_{1}+v_{2}+v_{3}) - 10(v_{1}-v_{2}) \}
     \{ -7(-v_{2}-v_{3}) + 3(v_{1}+v_{2}+v_{3}) - 10(v_{1}-v_{2}) \}
     \{ 8(-v_{2}-v_{3}) - 6(v_{1}+v_{2}+v_{3}) + 14(v_{1}-v_{2}) \}

     \ Image = < -4v_{1} + 20v_{2} + 10v_{3} ;  -7v_{1} + 20v_{2} + 10v_{3} ; -8v_{1} - 28v_{2} - 14v_{3} > butīs LD, so
     \ Image = \{ 10v_{1} + 20v_{2} + 1v_{3} ;  0v_{1} + 0v_{2} + 1v_{3} \}


    Unfortunately, both results are different!!
    which is the correct one? where is the mistake?
    Last edited by viko; April 18th 2010 at 09:30 AM. Reason: correcting mistake tonio noticed
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  2. #2
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    Quote Originally Posted by viko View Post
    Hi!

    i have a doubt here:

    Excercise:
    Given:
    B_{1}=\{v_{1},v_{2},v_{3}\}
    B_{2}=\{-v_{2}-v_{3},v_{1}+v_{2}+v_{3},v_{1}-v_{2}\}
    Both are basis from the V linear space.
    and
    f:V \rightarrow V \ / \ M(f)_{B_{1}B_{2}}= \left(\begin{array}{ccc}1&-2&1\\2&1&-2\\-1&-3&3\end{array}\right)

    -->Find Image(fof)

    So, my plan was:

    Get the change of basis matrix, this is easy because of the way the basis are expresed
     C_{B_{2}B_{1}} = \left(\begin{array}{ccc}0&0&1\\-1&1&-1\\-1&1&0\end{array}\right)



    Seems like you go by columns; fine, it's just the same...as long as you keep it being constant. Anyway, the second column's first entry (the matrix's 12 entry) must be 1, not zero.[/color]


    Now i know that C_{B_{2}B_{1}}.M(f)_{B_{1}B_{2}}=M(f)_{B_{1}B_{1}}

    \left(\begin{array}{ccc}0&0&1\\-1&1&-1\\-1&1&0\end{array}\right) \left(\begin{array}{ccc}1&-2&1\\2&1&-2\\-1&-3&3\end{array}\right) = \left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right)" alt="
    \left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right)" />

    Now i just do
    M(f)_{B_{1}B_{1}}M(f)_{B_{1}B_{1}}
    \Rightarrow \left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right) \left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right) = \left(\begin{array}{ccc}-2&-11&10\\8&14&-16\\4&7&-8\end{array}\right)

    And those columns are the image of  (fof) but expresed in B_{1}, so, image well expresed is
     \ Image = < -2v_{1} + 8 v_{2} + 4v_{3} ; -11v_{1} + 14v_{2} + 7v_{3} ; 10v_{1} -16v_{2} - 8v_{3} > butīs LD, so
     \ Image = \{ -2v_{1} + 8 v_{2} + 4v_{3} ; -11v_{1} + 14v_{2} + 7v_{3} \}
    -----------------------------------------

    So, the problem is that in the book i found a proposition that says that

    given linear functions  f:V \rightarrow W \ and  g:W \rightarrow U

    \Rightarrow \ M(gof)_{B_{1}B_{3}} = M(g)_{B_{2}B_{3}}.M(f)_{B_{1}B_{2}}

    -----------------------------------------------------

    So, that, what i have done is:  M(f)_{B_{1}B_{2}}.M(f)_{B_{1}B_{2}} = M(fof)_{B_{1}B_{2}} = \left(\begin{array}{ccc}-4&-7&8\\6&3&-6\\-10&-10&14\end{array}\right)


    [color=red]This is all you need: this is the correct answer.

    Tonio


    and i know now that the image of the LT are the colums of the matrix, but those are expresed in  B_{2} so, the image is going to be formed by:

     \{ -4(-v_{2}-v_{3}) + 6(v_{1}+v_{2}+v_{3}) - 10(v_{1}-v_{2}) \}
     \{ -7(-v_{2}-v_{3}) + 3(v_{1}+v_{2}+v_{3}) - 10(v_{1}-v_{2}) \}
     \{ 8(-v_{2}-v_{3}) - 6(v_{1}+v_{2}+v_{3}) + 14(v_{1}-v_{2}) \}

     \ Image = < -4v_{1} + 20v_{2} + 10v_{3} ; -7v_{1} + 20v_{2} + 10v_{3} ; -8v_{1} - 28v_{2} - 14v_{3} > butīs LD, so
     \ Image = \{ 10v_{1} + 20v_{2} + 1v_{3} ; 0v_{1} + 0v_{2} + 1v_{3} \}


    Unfortunately, both results are different!!
    which is the correct one? where is the mistake?
    .
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  3. #3
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    Oct 2009
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    Thanks in reply tonio
    Note that the mistake in the change of basis matrix was just when I copied the matrix to computer, the other numbers have been got with the correct matrix

    Anyway i still dont understand why are the results different if both procedures are correct! :S
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  4. #4
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    Joined
    Oct 2009
    Posts
    10
    Hi, i discovered the problem!

    Quote Originally Posted by myself
    -----------------------------------------

    So, the problem is that in the book i found a proposition that says that

    given linear functions  f:V \rightarrow W \ and  g:W \rightarrow U

    \Rightarrow \ M(gof)_{B_{1}B_{3}} = M(g)_{B_{2}B_{3}}.M(f)_{B_{1}B_{2}}

    -----------------------------------------------------

    So, that, what i have done is:  M(f)_{B_{1}B_{2}}.M(f)_{B_{1}B_{2}} = M(fof)_{B_{1}B_{2}} = \left(\begin{array}{ccc}-4&-7&8\\6&3&-6\\-10&-10&14\end{array}\right)
    Thats completely BAD because here:

     M(f)_{B_{1}B_{2}}.M(f)_{B_{1}B_{2}} = M(fof)_{B_{1}B_{2}} = \left(\begin{array}{ccc}-4&-7&8\\6&3&-6\\-10&-10&14\end{array}\right)

    basis 1 and 2 must be the same, and ARE NOT. So, before doing this, is mandatory to apply a change of basis.
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