1. 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

2. 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!)