QR Decomposition of Matrix: Solve for Answers

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Homework Help Overview

The discussion revolves around the QR decomposition of a given 2x2 matrix. The original poster attempts to decompose the matrix and is seeking assistance with the second part of the problem, specifically in determining the vector q2.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of QR decomposition and the normalization of vectors. The original poster shares their attempts and calculations, expressing confusion about the second vector q2.

Discussion Status

Some participants have provided guidance on the normalization process, suggesting that the vector q2 should have unit norm. The original poster has confirmed that they found the normalization helpful, indicating a productive direction in the discussion.

Contextual Notes

There is a mention of missing definitions for symbols used in the calculations, which some participants note may contribute to the confusion in following the original poster's reasoning.

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Homework Statement


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}

Homework Equations


[tex]c_2 = (v_2 * u_1)q_1 + \parallel w_2 \parallel q_2[/tex]

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}<br /> 1 & \\<br /> -1 &<br /> \end{bmatrix}[/tex]
[tex]\parallel w_2 \parallel = -49/25[/tex]
Therefore,
[tex]\begin{bmatrix}<br /> 1 & \\<br /> -1 &<br /> \end{bmatrix} = .2 \begin{bmatrix}<br /> .8 & \\<br /> .6 &<br /> \end{bmatrix} - 49/25q_2[/tex]
step two
[tex]\begin{bmatrix}<br /> 1 & \\<br /> -1 &<br /> \end{bmatrix} = \begin{bmatrix}<br /> 4/25 & \\<br /> 3/25 &<br /> \end{bmatrix} - 49/25q_2[/tex]
step three
[tex] \begin{bmatrix}<br /> (25-4)/25 & \\<br /> (-25-3)/25 &<br /> \end{bmatrix} = 49/25q_2[/tex]
step four
[tex] \begin{bmatrix}<br /> 21/49 & \\<br /> -28/49 &<br /> \end{bmatrix} = q_2[/tex]
The answer is supposed to be
[tex] \begin{bmatrix}<br /> .6 & \\<br /> -.8 &<br /> \end{bmatrix} = q_2[/tex]
So I made an error somewhere.
 
Last edited:
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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.
 
fzero said:
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.

what do you mean by properly normalize.
 
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.
 
amazing it worked! i was skeptical that it would but it did!
 

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