A = QR Decomposition

1. Jul 27, 2012

g.lemaitre

1. The problem statement, all variables and given/known data
Decompose the following matrix using QR decomposition
\begin{bmatrix}
4 & 1 \\
3 & -1
\end{bmatrix}
\begin{bmatrix}
.8 & .6 \\
.6 & .8
\end{bmatrix}
The following matrix is supposed to be next to the previous but I can't figure out how to do that. Any help in that area would be appreciated.
\begin{bmatrix}
5 & .2 \\
0 & 1.4
\end{bmatrix}
2. Relevant equations
$$c_2 = (v_2 * u_1)q_1 + \parallel w_2 \parallel q_2$$
3. The attempt at a solution
I was able to get the first part of the answer
\begin{bmatrix}
.8 & \\
.6 &
\end{bmatrix}
It's the second part
\begin{bmatrix}
.6 & \\
-.8 &
\end{bmatrix}
that i'm having trouble with. I'm also not worried about the R part right now.
Ok, let's plug the numbers into this equation:
$$c_2 = (v_2 * u_1)q_1 + \parallel w_2 \parallel q_2$$
$$v_2 * u_1 = .2$$
$$c_2 = \begin{bmatrix} 1 & \\ -1 & \end{bmatrix}$$
$$\parallel w_2 \parallel = -49/25$$
Therefore,
$$\begin{bmatrix} 1 & \\ -1 & \end{bmatrix} = .2 \begin{bmatrix} .8 & \\ .6 & \end{bmatrix} - 49/25q_2$$
step two
$$\begin{bmatrix} 1 & \\ -1 & \end{bmatrix} = \begin{bmatrix} 4/25 & \\ 3/25 & \end{bmatrix} - 49/25q_2$$
step three
$$\begin{bmatrix} (25-4)/25 & \\ (-25-3)/25 & \end{bmatrix} = 49/25q_2$$
step four
$$\begin{bmatrix} 21/49 & \\ -28/49 & \end{bmatrix} = q_2$$
The answer is supposed to be
$$\begin{bmatrix} .6 & \\ -.8 & \end{bmatrix} = q_2$$
So I made an error somewhere.

Last edited: Jul 27, 2012
2. Jul 27, 2012

fzero

It's tough to follow because you haven't defined every symbol, but $q_2$ should have unit norm. If you properly normalize your result, you'll find that it agrees with the solution.

3. Jul 27, 2012

g.lemaitre

what do you mean by properly normalize.

4. Jul 28, 2012

who_

To normalize a vector q is to find q / ||q||. Bascially, it means to scale the vector so that it lies on the unit sphere.

5. Jul 28, 2012

g.lemaitre

amazing it worked! i was skeptical that it would but it did!