Degenerate Eigenvalues and Eigenvectors: Understanding Differences in Solutions

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The discussion centers on the confusion regarding the eigenvalues and eigenvectors of a matrix, particularly in cases of degeneracy. It highlights that while the first method yields four eigenvalues, they are actually two distinct values that are double degenerate. The second method, which involves taking determinants of specific rows, also results in the same eigenvalues but allows for more flexibility in choosing eigenvectors due to the degeneracy. Participants clarify that the eigenvalues are not distinct in the degenerate case, leading to the conclusion that the two methods are not equivalent. The conversation emphasizes the importance of recognizing the nature of eigenvalues in determining the characteristics of eigenvectors.
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Homework Statement


Please see the attached image.

The first line just finds the eigenvalues of that matrix.

The second line finds the eigenvectors.

The third line just takes row 1 and row 3 of that matrix and find the determinant.
The fourth line just takes row 2 and row 4 of that matrix and find the determinant.


Because the two sets of equations are identitical, the eigenval
ues are double degenerate in the later case. Thus the evectors are not fixed.

But in the former case, the eigenvalues/eigenvecotrs are different.


THe solution is the later but I don't understand why the former part gives different answers.

What's wrong?

Homework Equations





The Attempt at a Solution

 

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I'm not sure what you're getting at. You're getting the same set of eigenvalues whether you use the 4x4 matrix or the two 2x2 matrices. Are you asking why you have more freedom to choose the eigenvectors in the 4x4 case?
 
No that's the thing. The eigenvalues are not the same.

Is the technique on line 3 and 4 valid? the equations are independent of each other.
 
You have the same set of four eigenvalues. How are the two sets different?
 
In the first line, there were 4 distinct eigenvalues.In the third & fourth line there are 2 degenerate eigenvalues.

Do you know what I mean?

Line 3 examines row 1 and row 3 in the matrix and takes the determinant of that seperately from

Line 4 which examines row 2 and row 4 in the matrix and takes the determinant of that.

Can you do this? Clearly, they're not equivalent.
 
Nusc said:
In the first line, there were 4 distinct eigenvalues.

How exactly do you consider -\sqrt{3}\sqrt{3A^2+4B^2}[/itex] to be distinct from -\sqrt{3}\sqrt{3A^2+4B^2}[/itex]? <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" /><br /> <br /> In your first line, you do not have 4 distinct eigenvalues; you have two doubly degenerate eigenvalues.
 
Those values each appear twice among the four eigenvalues in both cases. How are they different?
 
Haha you're right. I took the determinant by hand and solved it and didn't see what was written in mathematica...
 

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