Degenerate Eigenvalues

Homework Statement

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?

The Attempt at a Solution

Attachments

• Picture 2.jpg
41 KB · Views: 510

vela
Staff Emeritus
Homework Helper
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.

vela
Staff Emeritus
Homework Helper
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.

gabbagabbahey
Homework Helper
Gold Member
In the first line, there were 4 distinct eigenvalues.

How exactly do you consider [tex]-\sqrt{3}\sqrt{3A^2+4B^2}[/itex] to be distinct from [tex]-\sqrt{3}\sqrt{3A^2+4B^2}[/itex]?

In your first line, you do not have 4 distinct eigenvalues; you have two doubly degenerate eigenvalues.

vela
Staff Emeritus