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Math Help - Does the matrix span R^4?

  1. #1
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    Does the matrix span R^4?

    How can I determine if the matrix A spans R^4?

    A = \begin{bmatrix}-2 & -5 & 8 & 0 & -17\\ 1 & 3 & -5 & 1 & 5\\ 3 & 11 & -19 & 7 & 1\\ 1 & 7 & -13 & 5 & -3 \end{bmatrix}
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    Quote Originally Posted by posix_memalign View Post
    How can I determine if the matrix A spans R^4?
    Do you mean, "How can I determine if the columns of matrix A span \mathbb{R}^4?"

    A = \begin{bmatrix}-2 & -5 & 8 & 0 & -17\\ 1 & 3 & -5 & 1 & 5\\ 3 & 11 & -19 & 7 & 1\\ 1 & 7 & -13 & 5 & -3 \end{bmatrix}
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    Quote Originally Posted by Ackbeet View Post
    Do you mean, "How can I determine if the columns of matrix A span \mathbb{R}^4?"
    Yes, sorry about that. :-)
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    Right-ho. Then the way I would go about it is this: let the columns of A be denoted \mathbf{a}_{1},\mathbf{a}_{2},\dots,\mathbf{a}_{5}  . They are column vectors in \mathbb{R}^{4}. Let

    \mathbf{r}=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{  4}\end{bmatrix}

    be an arbitrary vector in \mathbb{R}^{4}. We want to know if there is a set of scalars, b_{1},b_{2},\dots,b_{5}, such that

    b_{1}\mathbf{a}_{1}+b_{2}\mathbf{a}_{2}+\dots+b_{5  }\mathbf{a}_{5}=\mathbf{r},

    right? Now, this equation that I've just written down can be translated into a system of equations for the unknown constants b_{j}. Your goal is to determine the rank of the coefficient matrix for that system. If the rank is 4, then you've got yourself a spanning set with those \mathbf{a}_{j}'s. Otherwise, if the rank is less than 4, you don't. Make sense? Where would you go next?
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    Quote Originally Posted by Ackbeet View Post
    Right-ho. Then the way I would go about it is this: let the columns of A be denoted \mathbf{a}_{1},\mathbf{a}_{2},\dots,\mathbf{a}_{5}  . They are column vectors in \mathbb{R}^{4}. Let

    \mathbf{r}=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{  4}\end{bmatrix}

    be an arbitrary vector in \mathbb{R}^{4}. We want to know if there is a set of scalars, b_{1},b_{2},\dots,b_{5}, such that

    b_{1}\mathbf{a}_{1}+b_{2}\mathbf{a}_{2}+\dots+b_{5  }\mathbf{a}_{5}=\mathbf{r},

    right? Now, this equation that I've just written down can be translated into a system of equations for the unknown constants b_{j}. Your goal is to determine the rank of the coefficient matrix for that system. If the rank is 4, then you've got yourself a spanning set with those \mathbf{a}_{j}'s. Otherwise, if the rank is less than 4, you don't. Make sense? Where would you go next?
    Thank you so much. :-) I would row reduce the matrix to row reduced echelon form and obtain that the first four columns are linearly independent -- i.e. the rank is 4. Is this correct at all?
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    Quote Originally Posted by posix_memalign View Post
    Thank you so much. :-) I would row reduce the matrix to row reduced echelon form and obtain that the first four columns are linearly independent -- i.e. the rank is 4. Is this correct at all?
    You would reduce? Or you did reduce?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    You would reduce? Or you did reduce?
    I did reduce, I had actually already done that when I was fumbling around without knowing if it was the correct thing to do, I got:

    \begin{bmatrix}1 & 0 & 0 & 0 & 27/10\\ 0 & 1 & 0 & 0 & 3/10\\ 0 & 0 & 1 & 0 & -17/10\\ 0 & 0 & 0 & 1 & -5 \end{bmatrix}

    Hence why the first four columns are linearly independent, thus the rank is 4.
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  8. #8
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    Hmm. WolframAlpha disagrees with you. Maybe you should re-check your work?
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  9. #9
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    Quote Originally Posted by Ackbeet View Post
    Hmm. WolframAlpha disagrees with you. Maybe you should re-check your work?
    Oh dear ...
    Well I redid it and see now that the rank is 3. I.e. it does not span R^4.
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  10. #10
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    That's correct, it does not.
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