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Math Help - [SOLVED] Basis of subspace R

  1. #1
    Member mybrohshi5's Avatar
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    [SOLVED] Basis of subspace R

    Find a basis of the subspace of consisiting of all vectors of the form

     \begin{bmatrix} x_1\\6x_1+x_2\\7x_1+3x_2\\-2x_1+6x_2\\ \end{bmatrix}

    i am not sure how to find this since there are variables  x_1 and  x_2

    Thank you for any help
    Last edited by mybrohshi5; March 28th 2010 at 01:15 PM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint :

    <br />
\begin{bmatrix} x_1\\6x_1+x_2\\7x_1+3x_2\\-2x_1+6x_2\\ \end{bmatrix} = x_1\begin{bmatrix} 1\\6\\7\\-2\\ \end{bmatrix} +x_2\begin{bmatrix} 0\\1\\3\\6\\ \end{bmatrix} <br />
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  3. #3
    Member mybrohshi5's Avatar
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    The basis are just the two you listed because when

     \begin{bmatrix}1&0&0 \\ 6&1&0 \\ 7&3&0 \\ -2&6&0 \\ \end{bmatrix}

    is in reduced row echelon form i got

     \begin{bmatrix}1&0& 0& 0 \\ 0& 1& 0& 0 \\ 0& 0& 0& 0& \\ 0& 0& 0& 0 \\ \end{bmatrix}

    which showed me that  \begin{bmatrix} 1\\6\\7\\-2\\ \end{bmatrix} and \begin{bmatrix} 0\\1\\3\\6\\ \end{bmatrix}

    are just the basis of the subspace R.

    Does that seem correct?

    Thanks for the hint
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Well it's clear from the hint I gave you that every vector in your subspace can be written as a linear combination of the two vectors above. So the two vectors form a basis if and only if they are linearly independent. It's very easy to check for linear independence when you have only two vectors; two vectors are linearly dependent if and only if one is a multiple of the other. It's clearly not the case here! Your method is also valid (though I'm not sure why you added a third zero column to the matrix you reduced.)
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  5. #5
    Member mybrohshi5's Avatar
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    Thank you that makes sense and actually helps me a lot about understanding these.

    One quick (and kind of a stupid question) but could you please quickly explain what you mean by "one is a multiple of the other."

    I added the extra columns of zero's cause when i plug the matrix into my calculator to get the rref it needed extra columns of zeros in order to calculate it without my calculator saying error

    Thanks again
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Well, for example, the vectors (1,1,1), (2,2,2) are linearly dependent because (2,2,2)=2(1,1,1). But the vectors (1,1,1),(2,2,3) are linearly independent because (1,1,1) \neq k(2,2,3) for all k \in \mathbb R. That's what I mean by one being a multiple of the other.
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  7. #7
    Member mybrohshi5's Avatar
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    Perfect thank you so much for the help and explaining that i feel silly cause i knew it was easy but i just couldnt remember what a multiple meant haha.
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