Results 1 to 13 of 13

Math Help - Matrix of linear transformation, basis problem

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Matrix of linear transformation, basis problem

    Let T be a linear transformation in \mathbb{R}^3 by T(x_1,x_2,x_3)=(x_1-x_3, x_1+3x_2,2x_2-x_3).
    1)Give the matrix of T with respect to the canonical basis \bold C.
    2)Tell whether T is invertible or no and if it is calculate T^{-1}.
    3)Give the matrix of T with respect to the basis \bold {B} =\{ (1,0,1),(-1,2,1),(2,1,1)\}.
    4)Give the matrix of T with respect to the basis \bold C, \bold B.
    5)Give the matrix of T with respect to the basis \bold B, \bold C.
    ------------------------------------------------
    My attempt :
    1) [T]_{\bold C} = \begin{bmatrix} 1&0&-1 \\ 1&3&0 \\ 0&2&-1   \end{bmatrix}.
    2)As [T]_{\bold C} is invertible, so is T. I calculated [T]_{\bold C}^{-1} which gave me T^{-1}(x_1,x_2,x_3)=( \frac{3x_1+2x_2-3x_3}{5}   , \frac{-x_1+x_2}{5}-\frac{x_3}{10}, \frac{-2x_1+2x_2-3x_3}{5}).
    3) I don't know how to do it! I guess I have to express the vectors of the basis of B as linear combination of the canonical vectors... but I'll have 3 scalars and I don't know what to do with them.
    4)I think they ask for [T]_{\bold B}^{\bold C}.
    I wrote \left[\begin{array}{ccc|ccc}     1 & -1 & 2 &1&0&-1 \\     0 & 2 & 1 & 1&3&0 \\     1 & 1 & 1 & 0&2&-1   \end{array}\right] and I reduced the left matrix. At last the right matrix is the one they ask for and I got it to be \begin{bmatrix} -1& -\frac{1}{2}& -1 \\ 0& \frac{5}{4}&0 \\ 1 & \frac{1}{2}&0 \end{bmatrix}.
    5) If my attempt for 4) is good, I know how to do this one.
    ------------------------------------------------------------------

    I really need help for part 3), and I'd like to know if I did well what I did. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    c)

     T \left(\begin{array}{ccc}1\\0\\1\end{array}\right)  = \left(\begin{array}{ccc}0\\1\\\text{-}1\end{array}\right)

     T \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right)  = \left(\begin{array}{ccc}\text{-}2\\5\\3\end{array}\right)

     T \left(\begin{array}{ccc}2\\1\\1\end{array}\right)  = \left(\begin{array}{ccc}1\\5\\1\end{array}\right)

    so the matrix of T with respect to basis B is

     \left(\begin{array}{ccc}0&\text{-}2&1\\1&5&5\\\text{-}1&3&1\end{array}\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thanks, you saved me (I understand what you did). You don't know how valuable is your help for me. Tomorrow is the final exam which lasts 4 hours and I just can't mess it up because I already messed up the previous one.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    For (d) I would do the following:

    T \left(\begin{array}{ccc}1\\0\\0\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\1\end{array}\right)  , \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right) , \left(\begin{array}{ccc}2\\1\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga  mma_{1}\end{array}\right)


    T \left(\begin{array}{ccc}0\\1\\0\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\1\end{array}\right)  , \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right) , \left(\begin{array}{ccc}2\\1\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{2}\\\beta_{2}\\\ga  mma_{2}\end{array}\right)

    T \left(\begin{array}{ccc}0\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\1\end{array}\right)  , \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right) , \left(\begin{array}{ccc}2\\1\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{3}\\\beta_{3}\\\ga  mma_{3}\end{array}\right)

    then the matrix representation of T with respect to basis C and B is

    \left(\begin{array}{ccc}\alpha_{1}&\alpha_{2}&\alp  ha_{3}\\\beta_{1}&\beta_{2}&\beta_{3}\\\gamma_{1}&  \gamma_{2}&\gamma_{3}\end{array}\right)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Random Variable View Post
    c)

     T \left(\begin{array}{ccc}1\\0\\1\end{array}\right)  = \left(\begin{array}{ccc}0\\1\\\text{-}1\end{array}\right)

     T \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right)  = \left(\begin{array}{ccc}\text{-}2\\5\\3\end{array}\right)

     T \left(\begin{array}{ccc}2\\1\\1\end{array}\right)  = \left(\begin{array}{ccc}1\\5\\1\end{array}\right)

    so the matrix of T with respect to basis B is

     \left(\begin{array}{ccc}0&\text{-}2&1\\1&5&5\\\text{-}1&3&1\end{array}\right)
    I just called a friend and he told me it's not right...
    He told me that what you did here is to find [T]_{\bold C}^{\bold B} while the exercise asks for [T]_{\bold B}^{\bold B}.
    He told me that to find the columns of [T]_{\bold B}^{\bold B}, I have to find a, b and c such that (0,1,-1)=a(1,0,1)+b(-1,2,1)+c(2,1,1) for the first column, then (-2,5,3)=d(1,0,1)+e(-1,2,1)+f(2,1,1) and (d,e,f) is the second column of the matrix they ask for, and so on.

    I'm completely confused. I understand both what you did, but I don't know who's right. I'm at a loss!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    First of all, why did I call part (3) "part (c)" and part (4) "part (d)"? I must be seeing things.

    Anyways, how did you do part (1)?

     T \left(\begin{array}{ccc}1\\0\\0\end{array}\right) = \left(\begin{array}{ccc}1\\1\\0\end{array}\right)

     T \left(\begin{array}{ccc}0\\1\\0\end{array}\right) = \left(\begin{array}{ccc}0\\3\\2\end{array}\right)

     T \left(\begin{array}{ccc}0\\0\\1\end{array}\right) = \left(\begin{array}{ccc}\text{-}1\\0\\\text{-}1\end{array}\right)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    To find the matrix representation of T with respect to bases B and C, I would do the reverse of what I did before:

     T \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga  mma_{1}\end{array}\right)

     T \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{2}\\\beta_{2}\\\ga  mma_{2}\end{array}\right)

     T \left(\begin{array}{ccc}2\\1\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{3}\\\beta_{3}\\\ga  mma_{3}\end{array}\right)
    Last edited by Random Variable; June 28th 2009 at 05:31 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Random Variable View Post
    First of all, why did I call part (3) "part (c)" and part (4) "part (d)"? I must be seeing things.

    Anyways, how did you do part (1)?

     T \left(\begin{array}{ccc}1\\0\\0\end{array}\right) = \left(\begin{array}{ccc}1\\1\\0\end{array}\right)

     T \left(\begin{array}{ccc}0\\1\\0\end{array}\right) = \left(\begin{array}{ccc}0\\3\\2\end{array}\right)

     T \left(\begin{array}{ccc}0\\0\\1\end{array}\right) = \left(\begin{array}{ccc}\text{-}1\\0\\\text{-}1\end{array}\right)
    I did the same thing as you did here.
    And thanks for T with respect to bases B and C... it helps me.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Wait! But
    \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[<br />
\left(\begin{array}{ccc}1\\0\\0\end{array}\right) ,<br />
\left(\begin{array}{ccc}0\\1\\0\end{array}\right) ,<br />
\left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg]<br />
\left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga  mma_{1}\end{array}\right) gives (\alpha _1,\beta _1, \gamma _1)=(1,0,1)... so it's like if I didn't do anything. (multiplying a vector with the identity matrix).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by arbolis View Post
    Wait! But
    \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) ,
    \left(\begin{array}{ccc}0\\1\\0\end{array}\right) ,
    \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg]
    \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga mma_{1}\end{array}\right)" alt="
    \left(\begin{array}{ccc}1\\0\\0\end{array}\right) ,
    \left(\begin{array}{ccc}0\\1\\0\end{array}\right) ,
    \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg]
    \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga mma_{1}\end{array}\right)" /> gives (\alpha _1,\beta _1, \gamma _1)=(1,0,1)... so it's like if I didn't do anything. (multiplying a vector with the identity matrix).
    I forgot the T.

     T \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga  mma_{1}\end{array}\right)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Random Variable View Post
    I forgot the T.

     T \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga  mma_{1}\end{array}\right)
    Oh yes, that's it! I get it now.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Random Variable View Post
    Indeed it helps. There's an error in the first page when they give [T]_{\beta}^{\gamma}. The first entry of the matrix is 1/3 while it should be -1/3.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Basis of kernel(T) where T is a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: November 16th 2011, 01:42 PM
  2. Proof for a basis of a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: February 8th 2010, 06:00 AM
  3. linear transformation ad standard basis
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 23rd 2009, 06:12 AM
  4. Find the basis of a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 13th 2009, 02:02 PM
  5. Help with proving for a basis and a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 1st 2008, 04:12 AM

Search Tags


/mathhelpforum @mathhelpforum