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Kernel and image

  1. Jan 26, 2009 #1
    1. The problem statement, all variables and given/known data

    38) Determine whether or not v1 = (-2,0,0,2) and v2 = (-2,2,2,0) are in the kernel of the linear transformation T:R^4 > R^3 given by T(x) = Ax where

    A = [1 2 -1 1;
    1 0 1 1;
    2 -4 6 2]

    39) Determine whether or not w1 = (1,3,1) or w2 = (-1,-1,-2) is in the image of the linear transformation given in question 38?

    2. Relevant equations

    3. The attempt at a solution

    I row-reduced it to rref then i let matrix = 0 and then solve for x1, x2, x3 and x4 . Which gave me x3 (-1,1,1,0) + x4 (-1,0,0,1) . v1 and v2 are not in that kernel i found but the answer states otherwise. is it because v1 and v2 are just scalar multiples of x3 and x4 ?For question 39 , i am stuck too. Please help.
    Last edited: Jan 26, 2009
  2. jcsd
  3. Jan 26, 2009 #2
    To determine whether or not given vectors are in the kernel of a linear transformation, simply take the product of the matrix which represents the linear transformation and the vector in question.

    What should this product be if the vector is in the kernel?
  4. Jan 26, 2009 #3
    The product should give me a zero vector. What about for the image, question 39 ?
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