Understanding Degenerate Eigenvalues and Vectors in Math Operations

[bon]

Homework Statement

please see attached

Homework Equations

The Attempt at a Solution

Ok so I've done A and have worked out eigenvalues and vectors of H and B

For H I get 4 possible eigenvectors (1,0,0) (0,1,1) (0,0,1) and (0,1,0) . The q is why does neither matrix uniquely specify its eigenvectors? I guess the answer is because the eigenvalues are degenrate (repeated)..but is there a clearer explanation as to why?

Thanks

 

[bon]

Also - i can't seem to find three common eigen vectors.. I can only find (1,0,0) and (0.1.1)

 

[bon]

anyoneeee?

 

[bon]

why can't i find three common eigenvectors?

 

[rwisz]

for providing the attached information. Degenerate eigenvalues and vectors occur when there are multiple eigenvectors associated with the same eigenvalue. In other words, the eigenvalues are repeated, resulting in multiple possible solutions for the eigenvectors. This can happen when the matrix is not diagonalizable, meaning it cannot be transformed into a diagonal matrix with distinct eigenvalues. In your case, the matrices H and B do not have distinct eigenvalues, leading to multiple possible eigenvectors. This can also occur when there are repeated factors in the characteristic equation, resulting in repeated roots. The reason why the matrices do not uniquely specify their eigenvectors is because the eigenvectors are not unique due to the degenerate eigenvalues. I hope this helps clarify the concept of degenerate eigenvalues and vectors in math operations.