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Math Help - Something more??

  1. #1
    Junior Member simplysparklers's Avatar
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    Something more??

    Hey guys!
    I have a question here where I'm supposed to get the kernel and image of a linear transformation. I think I must be missing something, coz I feel like I haven't done enough!

    Let T: R^3 \rightarrow R^2 be the linear transformation defined by T((x_{1},x_{2},x_{3}))=(x_{1}+x_{2},x_{2}-2x_{3})

    (a) Find the kernel of T
    (b) Show that (1,0) and (0,1) are in the range of T. Determine the range of T.

    For (a), I just let:
    x_{1}+x_{2}=0,and
    x_{2}-2x_{3}=0,
    and I get x_{1}=-2x_{3}.
    So is my answer: kerT= (-2x_{3}+x_{2},x_{2}+x_{1})
    I'm guessing this is a stupid question, and I'm missing out loads! Help would be appreciated!
    Thanks,
    SimplySparklers
    Last edited by simplysparklers; November 10th 2008 at 09:54 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by simplysparklers View Post
    Hey guys!
    I have a question here where I'm supposed to get the kernel and image of a linear transformation. I think I must be missing something, coz I feel like I haven't done enough!

    Let T: R^3 \rightarrow R^2 be the linear transformation defined by T((x_{1},x_{2},x_{3}))=(x_{1}+x_{2},x_{2}-2x_{3})

    (a) Find the kernel of T
    (b) Show that (1,0) and (0,1) are in the range of T. Determine the range of T.

    For (a), I just let:
    x_{1}+x_{2}=0,and
    x_{2}-2x_{3}=0,
    and I get x_{1}=-2x_{3}.
    So is my answer: kerT= (-2x_{3}+x_{2},x_{2}+x_{1})
    I'm guessing this is a stupid question, and I'm missing out loads! Help would be appreciated!
    Thanks,
    SimplySparklers
    The Kernel of T is the set of all x \in \mathbb{R}^3 such that Tx=\bold{0}, so solving the equations (that you have above) that this implies gives the Kernel is the set \{(x,-x,-x/2): \ x\in \mathbb{R} \}

    CB
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