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Basis of the kernel

  1. Apr 12, 2008 #1
    Find a basis for Ker T that contains S = [tex]\begin{pmatrix}
    1\\
    0\\
    1\\
    0\\
    \end{pmatrix}[/tex], [tex]\begin{pmatrix}
    0\\
    1\\
    0\\
    2\\
    \end{pmatrix}[/tex] where [tex]T : R^4 -> R^4[/tex] is defined by

    [tex]T\begin{pmatrix}
    1\\
    b\\
    c\\
    d\\
    \end{pmatrix} = \begin{pmatrix}
    a - b - c\\
    a - 2b + c\\
    0\\
    0\\
    \end{pmatrix}[/tex].

    Well, I have found a basis 'B' for Ker (T) to be B ={[tex]\begin{pmatrix}
    3\\
    2\\
    1\\
    0\\
    \end{pmatrix}[/tex], [tex]\begin{pmatrix}
    0\\
    0\\
    0\\
    1\\
    \end{pmatrix}[/tex]}.

    I noticed that the two sets of vectors S and B are linearly independent of one another. Does this mean that there is no basis for Ker(T) that contains S, as a basis must be a minimal spanning set? Or have I gone astray somewhere?
     
  2. jcsd
  3. Apr 12, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Perhaps you should reread the problem. "Find a basis for the kernel of T that includes <1, 0, 1, 0> and <0, 1, 0, 2>" makes no sense as it is easy to see that those two vectors are NOT in the kernel of T and so cannot be in any basis for that kernel.
     
  4. Apr 12, 2008 #3
    Alright, that is what I had thought but just wanted to verify it with someone. Thank you very much!
     
  5. Apr 13, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    For future reference: A problem that says "Find a basis for vector space V that include vectors v1 and v2" , assuming v1 and v2 are in V and are independent, means that you are to find a basis that includes those two vectors and possibly more.
     
    Last edited: Apr 14, 2008
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