Finding Eigenvalues and questions

  • Thread starter Thread starter flyingpig
  • Start date Start date
  • Tags Tags
    Eigenvalues
Click For Summary

Homework Help Overview

The discussion revolves around finding eigenvalues of a matrix, specifically addressing the eigenvalue equation Ax = λx. The original poster presents their attempts at solving the problem and seeks clarification on methods for determining eigenvalues without technology.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set up the eigenvalue equation and reduce the matrix to find eigenvalues. They question the terminology related to linear dependence and seek a faster method to reach a triangular matrix. Other participants discuss the determinant method and the implications of known eigenvalues.

Discussion Status

The discussion is ongoing, with participants exploring different methods for calculating eigenvalues. Some guidance has been provided regarding the determinant approach, while the original poster expresses a preference for a quicker method. There is no explicit consensus on the best approach yet.

Contextual Notes

The original poster indicates that the problem is from an exam paper, which imposes restrictions on the use of technology for solving the problem.

flyingpig
Messages
2,574
Reaction score
1

Homework Statement



http://img703.imageshack.us/img703/4489/unledzh.th.png

Uploaded with ImageShack.us


The Attempt at a Solution



a)

Ax = λx

Ax = x

Ax - x = 0

(A - I)x = 0


I set up my matrix

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP154819f61i5gh6d9fed500006aa20fbb6bgf59g1?MSPStoreType=image/gif&s=17&w=207&h=56

RowReduced it and I got

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP155119f61i5gh6d9fed50000387i4eb69hc6g67e?MSPStoreType=image/gif&s=17&w=80&h=56

Because this matrix is linearly dependent (can I say the "set is linearly dependent?" I have trouble with these names like "set", "systems"), 1 is an eigenvalue.

b) I know I could just do it the long way, but is there a faster way? Because this is not in triangular form
 
Last edited by a moderator:
Physics news on Phys.org
I'm not sure what you mean with the long way, but afaik the method is as follows:

Ax=λx ⇒ det(A-λI)=0

Calculating this will give you a 3rd order polynomial to solve.
Luckily you already know one eigenvalue (which is 1), this makes it easier to solve and in passing by also verifies that 1 is an eigenvalue.

Do you know how to do this?
 
No... unfortunately
 
flyingpig said:
No... unfortunately

Well, here's a wiki page that explains how to take the determinant of a 3x3 matrix: http://en.wikipedia.org/wiki/Determinant#3-by-3_matrices

And your equation is:
\det \begin{pmatrix}-\lambda &amp; -4 &amp; -6 \\<br /> -1 &amp; -\lambda &amp; -3 \\<br /> 1 &amp; 2 &amp; 5-\lambda<br /> \end{pmatrix} = 0

Can you combine those to yield a 3rd order polynomial?

Btw, which tools are you allowed to use?
Are you allowed to use e.g. WolframAlpha?
 
No I know I could that, but this is a question from an exam paper, so no technologies. I am wondering if there is a faster way to get to a triangular matrix and just "read it off"
 
flyingpig said:
No I know I could that, but this is a question from an exam paper, so no technologies. I am wondering if there is a faster way to get to a triangular matrix and just "read it off"

Nope, I'm afraid not. This is the fastest way.
But you don't need to go to a triangular matrix to do this.
So though your answer to a) was correct (and yes, you can say that the set of equations is linearly dependent), doing it this way, would answer a) as well.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K