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Math Help - Finding a Linear Transformation given its Kernel

  1. #1
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    Finding a Linear Transformation given its Kernel

    Question: Give an example of a linear transformation whose kernel is the plane x + 2y + 3z = 0 in |R3.

    I'm lost here. I found the vectors [1, 1, -1] and [-5, 4, -1] span the Kernel, but I really have no idea where to go from here... would it be the product of the matrices that have kernels in the two lines in the direction of the two vectors?
    Last edited by Lord Voldemort; February 2nd 2011 at 09:13 AM.
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    Quote Originally Posted by Lord Voldemort View Post
    Question: Give an example of a linear transformation whose kernel is the plane x + 2y + 3x = 0 in |R3.

    I'm lost here. I found the vectors [1, 1, -1] and [-5, 4, -1] span the Kernel, but I really have no idea where to go from here... would it be the product of the matrices that have kernels in the two lines in the direction of the two vectors?
    You're in the right track! Now just define a lin. trans. T s.t. T(1,1,-1)=T(-5,4,-1)=(0,0,0) . For this, complete these two vectors to a basis of \mathbb{R}^3 and then define the l.t. on the third vector to be anything but the zero vector, and extend by linearity.

    Tonio

    Pd. I'm assuming you meant the plane x +2y +3z = 0 ...
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    MHF Contributor FernandoRevilla's Avatar
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    Choose, for example, the basis of \mathbb{R}^3:

    B=\{(1,1,-1),(-5,4,-1),(0,0,1)\}

    Then,

    T \equiv\begin{Bmatrix}T(1,1,-1)=(0,0,0)\\T(-5,4,-1)=(0,0,0)\\ T(0,0,1)=(a,b,c)\end{matrix}

    such that (a,b,c)\neq (0,0,0) determines a family of linear transformations with the given kernel.


    Fernando Revilla

    Edited: Sorry, I didn`t see tonio's post.
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    Sorry, we haven't covered the basis of a matrix/space, so completing a vector to a basis of |R3 doesn't mean anything yet.
    Last edited by Lord Voldemort; February 2nd 2011 at 09:08 AM.
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    Ok maybe not? I really don't know what you mean by completing something to a basis. Also as you said, "define the l.t. on the third vector to be anything but the zero vector", I don't understand how defining a new LT will do anything.
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Lord Voldemort View Post
    Ok maybe not? I really don't know what you mean by completing something to a basis. Also as you said, "define the l.t. on the third vector to be anything but the zero vector", I don't understand how defining a new LT will do anything.

    If you haven't covered the concept of matrix of a linear transformation with respect to determined basis, it is rather difficult to solve the above doubt.

    There are alternatives to solve your problem, but we need to know what tools we can use.


    Fernando Revilla
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  7. #7
    Senior Member Tinyboss's Avatar
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    How about f(x,y,z)=(x+2y+3z,0,0)?
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Tinyboss View Post
    How about f(x,y,z)=(x+2y+3z,0,0)?

    Right. That is good "auto content" solution. Also the family:


    f_{ (\alpha,\beta,\gamma)}(x,y,z)=(\alpha(x+2y+3z),\be  ta(x+2y+3z),\gamma(x+2y+3z))

    with  (\alpha,\beta,\gamma)\neq (0,0,0) .


    Fernando Revilla
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