# qr factorization

• Aug 23rd 2010, 05:32 PM
james12
qr factorization
$A =\left(
\begin{array}{cc}
1 & 3 \\
2 & 4 \\
-1 & -1 \\
0 & 1
\end{array}
\right)
$

$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 $R$ 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
• Aug 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!)