qr factorization

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August 23rd 2010, 05:32 PM
james12

qr factorization

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

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

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

Now I'm pretty sure that $R=Q^TA$ could someone please verify this and show what they got for $R$ thanks
August 23rd 2010, 05:46 PM
Ackbeet

Your candidate for $R$ will certainly imply that the equation $A=QR$ holds. The columns being orthonormal (as they are, by inspection), implies that $QQ^{T}=I$ , and $Q^{T}Q=I.$ Those are differently sized identity matrices, of course.

What I don't know off-hand is whether your candidate $R$ 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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