A = QR Decomposition

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}
the answer is
\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
[tex] c_2 = (v_2 * u_1)q_1 + \parallel w_2 \parallel q_2 [/tex]
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:
[tex] c_2 = (v_2 * u_1)q_1 + \parallel w_2 \parallel q_2 [/tex]
[tex]v_2 * u_1 = .2 [/tex]
[tex] c_2 = \begin{bmatrix}
1 & \\
-1 &
\end{bmatrix}
[/tex]
[tex] \parallel w_2 \parallel = -49/25 [/tex]
Therefore,
[tex] \begin{bmatrix}
1 & \\
-1 &
\end{bmatrix} = .2 \begin{bmatrix}
.8 & \\
.6 &
\end{bmatrix} - 49/25q_2 [/tex]
step two
[tex] \begin{bmatrix}
1 & \\
-1 &
\end{bmatrix} = \begin{bmatrix}
4/25 & \\
3/25 &
\end{bmatrix} - 49/25q_2 [/tex]
step three
[tex]
\begin{bmatrix}
(25-4)/25 & \\
(-25-3)/25 &
\end{bmatrix} = 49/25q_2[/tex]
step four
[tex]
\begin{bmatrix}
21/49 & \\
-28/49 &
\end{bmatrix} = q_2[/tex]
The answer is supposed to be
[tex]
\begin{bmatrix}
.6 & \\
-.8 &
\end{bmatrix} = q_2
[/tex]
So I made an error somewhere.

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.

g.lemaitre

what do you mean by properly normalize.

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.

g.lemaitre

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