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Math Help - Do the vectors (1,3,-2),(2,1,1) belong to R(T)?

  1. #1
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    Do the vectors (1,3,-2),(2,1,1) belong to R(T)?

    Let T:IR^3-->IR^3 be defined by T(a,b,c) = (a+b, b-2c, a + 2c). Do the vectors (1,3,-2), (2,1,1) belong to R(T)?

    Looking through my notes I found some clues to this question. I found that "RT is always span (Tv1,..,Tvn) where v1,..,vn is basis of domain V.

    My first approach was to make a column vector (a) = (a+b)
    (b) (b-2c)
    (c) (a+2c)

    and then remove the coefficients like this:

    a (1) b(1) c(0) to get a basis. I don't know if I was on the right track.
    (0) (1) (-2)
    (1) (0) (2)


    The second thing I tried was to compute two vectors for R(T) like this:

    T(1,3,-2) = (4,7,-3)

    T(2,1,1) = (3,-1,4)

    And then get that R(T) = span {(4,7,-3), (3,-1,4)}.

    Anyways, I didn't finish my answer since I am confused. Also sorry about my notation because I don't know what syntax I need to use to make a matrix on this forum.
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  2. #2
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    Quote Originally Posted by Undefdisfigure View Post
    Let T:IR^3-->IR^3 be defined by T(a,b,c) = (a+b, b-2c, a + 2c). Do the vectors (1,3,-2), (2,1,1) belong to R(T)?

    Looking through my notes I found some clues to this question. I found that "RT is always span (Tv1,..,Tvn) where v1,..,vn is basis of domain V.

    My first approach was to make a column vector (a) = (a+b)
    (b) (b-2c)
    (c) (a+2c)

    and then remove the coefficients like this:

    a (1) b(1) c(0) to get a basis. I don't know if I was on the right track.
    (0) (1) (-2)
    (1) (0) (2)


    The second thing I tried was to compute two vectors for R(T) like this:

    T(1,3,-2) = (4,7,-3)

    T(2,1,1) = (3,-1,4)

    And then get that R(T) = span {(4,7,-3), (3,-1,4)}.

    Anyways, I didn't finish my answer since I am confused. Also sorry about my notation because I don't know what syntax I need to use to make a matrix on this forum.

    Since T\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatri  x}a+b\\b-2c\\a+2c\end{pmatrix}, you only have to ask whether there exist a,b,c,\in\mathbb{R}\,\,\,s.t.\,\,\,\begin{pmatrix}  a+b\\b-2c\\a+2c\end{pmatrix}=\begin{pmatrix}\;\;1\\\;\;3\  \-2\end{pmatrix}\,\,or\,\,\begin{pmatrix}2\\1\\1\end  {pmatrix}.

    For example, for the first one, we'd get:

    ** a+b=1

    ** b-2c=3

    ** a+2c=-2

    Substracting the third eq. from the first one we get  b-2c=3= second eq., so the system seems to be congruent (i.e., has at least one solution), for example: a=2,b=-1,c=-2, so this first vector does belong to the range of T. Now you continue with the other one.

    Tonio
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    how do I thank you for this reply??
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  4. #4
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    Quote Originally Posted by Undefdisfigure View Post
    how do I thank you for this reply??

    You just have...any time!

    Tonio
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