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Math Help - QR-factorization

  1. #1
    Senior Member Twig's Avatar
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    QR-factorization

    Hi

    If  A = QR \mbox{ where } Q \mbox{ is } m \times n \mbox{ and } R n\times n , show that if the columns of A are linearly independent, then R must be invertible.
    (Hint: Study the equation  R\vec{x} = \vec{0} , and use the fact that  A = QR .

    Need some guidance here.
    I tried something like:

     R\vec{x} = \vec{0} \Rightarrow (Q^{T}A)\vec{x}=\vec{0}

    But  Q = \left[ \begin{matrix} \vec{u_{1}} \, \; ... \, \; \vec{u_{n}} \end{matrix} \right]
    And the columns of Q are orthonormal.

    So the matrix  (Q^{T}A) will have columns formed from the columns of  Q^{T} , using the weights in columns of A as weights.
    Is it correct to say that even the columns of  Q^{T} will be linearly independent?

    If so,  R = Q^{T}A will have linearly independent columns, and since R is n x n , R is invertible by the Invertible matrix theorem.

    Thanks!
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  2. #2
    Senior Member Twig's Avatar
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