
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 GramSchmidt 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

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 offhand 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!)