Linear Algebra - Find Orthogonal Matrix Q that diagonals

Click For Summary

Homework Help Overview

The discussion revolves around finding an orthogonal matrix Q related to a given matrix A in the context of linear algebra. Participants are exploring the normalization of vectors and the origin of a specific constant, 1/3, mentioned in an answer key.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the normalization process for vectors and how the constant 1/3 is derived. Questions are raised about which vector is being normalized and the method for calculating normalization constants.

Discussion Status

There is an ongoing exploration of the normalization concept, with some participants providing insights into the requirements for orthonormal vectors. Guidance has been offered regarding the normalization process, but there is no explicit consensus on the specific calculations or assumptions involved.

Contextual Notes

Participants are discussing the implications of normalization when vectors do not sum to the same value, indicating a potential complexity in the problem setup. The original poster expresses confusion regarding the answer key's constant and its derivation.

YoshiMoshi
Messages
233
Reaction score
10

Homework Statement


I'm told to find the matrix Q of the matrix A
equation 1.PNG


Homework Equations

The Attempt at a Solution



So my problem is that in the answer key they have S = (1/3)... and I have no idea where this 1/3 comes from. I get an equivalent answer for X_1, X_2, and X_3
S = [X_1, X_2, X_3] but when I form this matrix with the values for X_1, X_2, X_3 I'm not able to factor out a 1/3. Does anyone know where this value comes from?
This is the solutions

equation2.PNG

and this is my attempt, again I would get the same answer but have no idea were the 1/3 came from.
IMG_20160424_204918869.jpg

thanks for any help
 
Physics news on Phys.org
An orthogonal matrix is a unitary matrix. Therefore ##Q^*Q = QQ^*=I##. This is why the vectors making up the column of ##Q## must be orthonormal - 1/3 is a normalization constant.
 
How do I calculate this normalization constant? Thanks for your help. Like I'm exactly sure which vector I'm normalizing and getting 1/3. But I know to normalize a vector it's

V/||V||

Thanks for your help!
 
A vector is said to be normalized if its norm equal unity.
 
So it's just

1/sqrt((1+1+1)^2)?
 
YoshiMoshi said:
1/sqrt((1+1+1)^2)?
I don't understand what you want to say there.
Suppose you have three component vectors ##u = (a,b,c)^T##. In order to make it normalized you have to add a common constant by yourself to the vector: ##u = \gamma (a,b,c)^T##. By requiring its norm to be equal to one, namely ##\gamma^2(|a|^2 + |b|^2 + |c|^2) = 1##. This way you can find ##\gamma## in terms of the components of ##u##. For a general unitary matrix, ##\gamma## can be complex but since here the problem asks for an orthogonal matrix, ##\gamma## must be real.
 
Oh I see that makes more since to me know. So it's for the whole Matrix S? When I square all the terms in S and sum each row individually I get 9, so the constant must be 1/sqrt(9) = 1/3.

What if each row doesn't sum up to the same value like it did in this case?

Thanks for your help!
 
Then you cannot pull the common normalization constant out of the matrix's bracket. Nothing wrong with that.
 
ah ok thanks for your help
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K