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Math Help - Basis/Spans

  1. #1
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    Basis/Spans

    Let v_1=\begin{bmatrix}1  \\ 1 \\ a\end{bmatrix},v_2=\begin{bmatrix}1  \\ a \\ 1\end{bmatrix}, and v_3=\begin{bmatrix}a \\ 1 \\ 1\end{bmatrix}

    For which values of the parameter a is v_1,  v_2 and v_3 a basis for R^3?
    Last edited by Zocken; November 4th 2009 at 05:06 PM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint : evaluate the determinant of the matrix whose columns are those three vectors. Solve for the values of a for which this determinant is nonzero. (Do you see why?)
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  3. #3
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    Bruno, I found the determinant of the matrix to be a^3-3a+2 and the values of a that make it nonzero are a=/-2 and a=/1

    Is this right? And could you explain why this works (I am used to row reducing from class)? Thanks
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    How can you possibly get a^3 in there? You should get a second degree polynomial in a.

    This works, because for the three vectors to be a basis of \mathbb{R}^3 they should form the sides of a parallelepiped having nonzero volume, and the determinant is that volume. Or, more abstractly put, if \beta=\{v_1,v_2,v_3\} is a basis, then the change of basis transformation from \beta to the standard basis is invertible; its matrix representation is the one whose determinant you evaluated (with a small mistake).
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  5. #5
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    Determinant is easy method. But when dimension is high, it's difficult to solve equation.

    I suggest processing by rref

    rref(A)=rref[\vec{v}_1\ \vec{v}_2\ \vec{v}_3]=\begin{bmatrix} 1&1&a\\0&a-1&1-a\\0&0&2-a-a^2 \end{bmatrix}

    if three vectors span a basis, then rank(A)=3, a-1\neq 0 \& 2-a-a^2\neq 0
    Last edited by math2009; November 4th 2009 at 05:24 PM.
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  6. #6
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    Span

    thank you
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