Finding an eigenvector of 3x3 matrix

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Discussion Overview

The discussion revolves around finding an eigenvector of a 3x3 matrix, specifically focusing on the case where the eigenvalue λ is assumed to be 1. Participants explore the calculations involved in deriving the eigenvector and the implications of the matrix being a rotation matrix.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their calculations for the eigenvector based on the matrix (A - λI) and derives a vector t = k[-2.23, -4.02, 1], but finds a discrepancy with an online solver's result of (-0.016, 0.206, 0.978).
  • Another participant questions the assumption that λ = 1 and suggests verifying if the derived vector satisfies the eigenvector equation (At = λt).
  • A participant clarifies that λ = 1 is assumed because the matrix is a rotation matrix, which leads to the search for equivalent axes eigenvectors.
  • One participant acknowledges checking the eigenvalues and confirming that λ = 1 is indeed one of them.
  • A participant corrects a typo in their matrix and recalculates, obtaining a new eigenvector of (-0.0137, 0.225, 1), which is still different from the expected answer of (-0.0088, 0.216, 1).
  • Another participant suggests using more precision in calculations due to the nature of rotation matrices, which typically involve sine and cosine values.
  • One participant expresses uncertainty about their method for computing eigenvectors and seeks validation of their approach, reiterating their calculations and results.
  • A later reply indicates that the multiplication of (A - I) with the proposed eigenvector yields a result that is close to a zero vector, suggesting that the method may be correct but could suffer from rounding errors.
  • Finally, a participant confirms that using more precise values leads to results much closer to a zero vector, indicating that the method is likely correct despite initial discrepancies.

Areas of Agreement / Disagreement

Participants generally agree on the method of finding the eigenvector but express differing views on the accuracy of the results due to rounding errors and the precision of calculations. There is no consensus on the exact eigenvector values, as discrepancies remain unresolved.

Contextual Notes

Limitations include potential rounding errors in calculations and the dependence on the precision of the values used in the matrix. The discussion does not resolve the exact eigenvector values due to these uncertainties.

enc08
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Hi,

I'm trying to find an eigenvector of a matrix. I know that λ = 1, so my matrix (A - λI) is
[-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236]

And from rows 2 and 3 I get these simultaneous equations

-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0
0.1817t_{1}+0.1161t_{2}+0.0236t_{3}=0

I eliminate to find t_{2} = -4.02t_{3} and t_{1}=-2.23t_{3}

Thus the eigenvector is

t=k [-2.23, -4.02, 1]

But using an online solver gives the eigenvector as (-0.016, 0.206, 0.978).

Thanks for any pointers.
 
Last edited:
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Before you do anything else, check to see if it is an eigenvector (At = λt ?).
Also how do know λ = 1?
 
Hi,

The reason I'm assuming λ = 1 is because A is a rotation matrix (I didn't mention this in the original post), and I'm looking for the equivalent axes eigenvector.
 
As a sanity check, I have checked the eigenvalues and λ = 1 is one of them.
 
I realized I typo in my original matrix (matrix term 3,3 in the first post should be negative). I've correct it.

I get a closer answer, but my first term appears to be incorrect:

My answer: (-0.0137, 0.225, 1)
Given answer: (-0.0088, 0.216, 1)
 
Since your matrix is a rotation matrix, the entries are sin and cos values. I would try using more precision than the 4 decimal places you show, and see if that makes a difference in your resulting eigenvector.
 
I've just been given the numbers as is, and told that it's a rotation matrix.

Can you tell if my method for computing the eigenvectors is correct?

My approach is

λ = 1, so (A - λI) is
[-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236]

From rows 2 and 3:

-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0

0.1817t_{1}+0.1161t_{2}-0.0236t_{3}=0

Eliminate to find t_{2}=0.225t_{3} and t_{1}=-0.0137t_{3}

Thus the eigenvector is

t=k (-0.0137, 0.225, 1)

But the actual answer is given as (-0.0088, 0.216, 1).
 
enc08 said:
I've just been given the numbers as is, and told that it's a rotation matrix.

Can you tell if my method for computing the eigenvectors is correct?

My approach is

λ = 1, so (A - λI) is
[-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236]
Here's your matrix in a nicer form. To see what I did, right-click the matrix to view the underlying script.
$$A - I =\begin{bmatrix} -0.5253 & 0.8593 & -0.1906 \\ -0.8612 & -0.5018 & 0.1010 \\ 0.1817 & 0.1161 & -0.0236\end{bmatrix}$$

To check your work, I multiplied A - I above times your eigenvector x. That multiplication should produce a zero vector, but what I got was off by a little. My result was approximately < .0099, .0088, .00003>.

When you were row-reducing A - I, if you weren't careful with your arithmetic, you could have introduced a certain amount of imprecision in your results.


enc08 said:
From rows 2 and 3:

-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0

0.1817t_{1}+0.1161t_{2}-0.0236t_{3}=0

Eliminate to find t_{2}=0.225t_{3} and t_{1}=-0.0137t_{3}

Thus the eigenvector is

t=k (-0.0137, 0.225, 1)

But the actual answer is given as (-0.0088, 0.216, 1).
This is easy to check. Just multiply (A - I) times <-0.0088, 0.216, 1>.
 
Thanks Mark44. I have used more precise values, and using your technique for testing the eigenvector, I get an answer much closer to a zero vector.

That you've confirmed my method is correct will do, especially since the discrepancies seem only due to rounding errors. I would have much preferred the question to use round numbers, but never mind :)
 

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