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Math Help - Vectors spanning kernel of a matrix (Need guidance)

  1. #1
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    Vectors spanning kernel of a matrix (Need guidance)

    Question Details:
    Good day to all.
    I have just read the section in my textbook on rank, kernel and linear dependence/independence. I thought I understood the chapter but as I was doing the end of section exercises, I realized that there is something I have missed. The question that has left me wondering is as follows:

    Question:

    Find a matrix whose kernel is spanned by the vectors u = (1,3,2) and v = (-2,0,4)


    Does this mean that some linear combination of u and v say w = au + bv (a, b are real numbers) will produce Aw = 0 ??

    I am not sure it is relevant, but the topics of image and basis have not yet been covered.

    Any help would be greatly appreciated!
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Do you know what the span is? If you don't, you need to read on a bit more before you can answer this question.

    If you do, then you can definitely solve the question; the question is asking for a matrix M such that \mbox{ker }M = \mbox{span }\{u,v\}. Hint : fill up the matrix with variables, and obtain equations which need to be satisfied by the coefficients of the matrix for the statement to hold. You should be able not only to obtain one such matrix M, but a whole parametrized family of them.
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    Will look at that now! Thanks again.
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  4. #4
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    Quote Originally Posted by gate13 View Post
    Question Details:
    Good day to all.
    I have just read the section in my textbook on rank, kernel and linear dependence/independence. I thought I understood the chapter but as I was doing the end of section exercises, I realized that there is something I have missed. The question that has left me wondering is as follows:

    Question:

    Find a matrix whose kernel is spanned by the vectors u = (1,3,2) and v = (-2,0,4)


    Does this mean that some linear combination of u and v say w = au + bv (a, b are real numbers) will produce Aw = 0 ??
    It means that A(au+bv)= 0 for all a and b. But that will be true for any A such that Au= 0 and Av= 0. I recommend just writing A as
    \begin{bmatrix}p & q & r \\ s & t & u\\ v & w & x\end{bmatrix} and looking at \begin{bmatrix}p & q & r \\ s & t & u\\ v & w & x\end{bmatrix}\begin{bmatrix}1 \\ 3  \\ 2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix} and \begin{bmatrix}p & q & r \\ s & t & u\\ v & w & x\end{bmatrix}\begin{bmatrix}-2 \\ 0  \\ 4\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
    That will give you 6 equations for the 9 values in A so there are many correct answers.

    I am not sure it is relevant, but the topics of image and basis have not yet been covered.

    Any help would be greatly appreciated!
    Last edited by HallsofIvy; January 27th 2010 at 04:49 AM.
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  5. #5
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    Thanks to you too HallsofIvy for the detailed response. I have since begun extensive revision on the sections concerning rank, kernel and spans. I am going to start working on the question right away.
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