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Thread: qr factorization

  1. #1
    Junior Member
    Joined
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    qr factorization

    $\displaystyle A =\left(
    \begin{array}{cc}
    1 & 3 \\
    2 & 4 \\
    -1 & -1 \\
    0 & 1
    \end{array}
    \right)
    $

    $\displaystyle Q = \left(
    \begin{array}{cc}
    1/\sqrt{6} & 1/\sqrt{3}\\
    2/\sqrt{6} & 0\\
    -1/\sqrt{6} & 1/\sqrt{3}\\
    0 & 1/\sqrt{3}
    \end{array}
    \right)
    $

    The columns of Q were obtained by applying the Gram-Schmidt Process to the columns of A. Find the upper triangular matrix $\displaystyle R$ such that $\displaystyle A=QR$

    Now I'm pretty sure that $\displaystyle R=Q^TA$ could someone please verify this and show what they got for $\displaystyle R$ thanks
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  2. #2
    A Plied Mathematician
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    Your candidate for $\displaystyle R$ will certainly imply that the equation $\displaystyle A=QR$ holds. The columns being orthonormal (as they are, by inspection), implies that $\displaystyle QQ^{T}=I$, and $\displaystyle Q^{T}Q=I.$ Those are differently sized identity matrices, of course.

    What I don't know off-hand is whether your candidate $\displaystyle R$ is upper triangular. Why don't you post your calculations, and I'll verify those? (That's more in the spirit of this forum anyway!)
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