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Math Help - Nullity of a transformation

  1. #1
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    Nullity of a transformation

    T(x,y) = ((2x-(y/2)), -x+3y, 2y)

    I have found the kernel which is 0 how do we find the nullity of this?
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  2. #2
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    Quote Originally Posted by adam_leeds View Post
    T(x,y) = ((2x-(y/2)), -x+3y, 2y)

    I have found the kernel which is 0 how do we find the nullity of this?
    nullity is the dimension of kernel(T),..

    so, dim(ker(T)) = . . .
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  3. #3
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    Do you know what the dimension of the "trivial" subspace, the one containing only the 0 vector, is?
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  4. #4
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    Quote Originally Posted by dedust View Post
    nullity is the dimension of kernel(T),..

    so, dim(ker(T)) = . . .
    0?
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  5. #5
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    Quote Originally Posted by adam_leeds View Post
    0?
    yes,..
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  6. #6
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    Quote Originally Posted by dedust View Post
    yes,..
    if the kernel was x = 2 y = 0, would the nullity be 1?
    Last edited by adam_leeds; December 27th 2009 at 08:03 AM.
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  7. #7
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    Quote Originally Posted by adam_leeds View Post
    if the kernel was x = 2 y = 0, would the nullity be 0?
    you mean the kernel is \left({\begin{matrix}<br />
  2\\<br />
0\\<br />
\end{matrix}}\right)?
    what is the dimension of the space that spanned by this vector?
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  8. #8
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    Quote Originally Posted by dedust View Post
    you mean the kernel is \left({\begin{matrix}<br />
2\\<br />
0\\<br />
\end{matrix}}\right)?
    what is the dimension of the space that spanned by this vector?
    Sorry ill re phrase

    to find the kernel of a transformation

    you make the equations = 0

    and solve for x and y

    if 1 of x and y doesnt = 0

    is the dimension 1?

    and if they are both non zero solutions then the dimension is 2?
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  9. #9
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    Quote Originally Posted by adam_leeds View Post
    if the kernel was x = 2 y = 0, would the nullity be 1?
    Run the vector (2,0) through your transformation. What do you get?
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  10. #10
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    Quote Originally Posted by ands! View Post
    run the vector (2,0) through your transformation. What do you get?
    4, -2, 0
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  11. #11
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    Quote Originally Posted by adam_leeds View Post
    4, -2, 0
    Therefore it is NOT in the kernel of the linear transformation.

    But, since you said "I have found the kernel which is 0" in your first post, I think you were asking in general, not for this particular transformation. It is not possible for the kernel of a linear transformation to be a single non-zero vector!

    The kernel of a linear transformation, A, is the set of all vectors, v, such that Av= 0. That is not true for (2, 0). Further, if the kernel is not "trivial", that is, if Av= 0 for any non-zero v, it cannot consist of a single vector! If v is a non-zero vector such that Av= 0 then A(xv)= xAv= 0 for any number x so there must exist an infinite number of vectors in the kernel. The kernel of a linear transformation is a subspace, possibly trivial, which is why it has a dimension.
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