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Uniqueness of Degenerate Eigenfunctions

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Say you have some matrix and the eigenvalues are all the same (degenerate). Let’s say you have $N$ eigenvectors with the label $n$, call them $psi_n$. If the eigenvalues were not degenerate, then the eigenvectors would form a unique orthonormal basis for some space $mathbbV_N$. However, since they are degenerate, then there is no unique basis because they all have the same eigenvalues, so linear combinations of the eigenvectors are also eigenvectors. My question is:

if we choose any arbitrary basis, will the eigenvectors be orthogonal to each other?

For example, I can choose one eigenvector to be $psi_n$ and another $alphapsi_n + betapsi_n+1$, and these are already not orthogonal.

linear-algebra matrices eigenvalues-eigenvectors

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edited yesterday

YuiTo Cheng

2,0192636

asked yesterday

Josh PilipovskyJosh Pilipovsky

1011

New contributor

Josh Pilipovsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand your question, and your second sentence is not true. Choose an arbitrary basis of what? What does this basis have to do with your eigenvectors? Why isn't the last sentence of your question already an answer?
  $endgroup$
  – Qiaochu Yuan
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A matrix all of whose eigenvalues are the same number $lambda$ necessarily has the form $lambda I + N$ where $N$ is a nilpotent matrix. If it is $n times n$, then it has $n$ linearly independent eigenvectors iff $N = 0$ iff the matrix is just a scalar multiple of the identity, so any nonzero vector is an eigenvector.
  $endgroup$
  – Qiaochu Yuan
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is a "degenerate" eigenvalue?
  $endgroup$
  – David C. Ullrich
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Look, if you have all the eigenvalues the same for a matrix, then the eigenvectors are not unique. You can form linear combinations of them, and they will still be eigenvectors of the original matrix. What I'm asking is when you have already found all the eigenvectors, i) will any linear combination of them form a viable basis?, and ii) will this basis necessarily be orthonormal?
  $endgroup$
  – Josh Pilipovsky
  18 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Say you have some matrix and the eigenvalues are all the same (degenerate). Let’s say you have $N$ eigenvectors with the label $n$, call them $psi_n$. If the eigenvalues were not degenerate, then the eigenvectors would form a unique orthonormal basis for some space $mathbbV_N$. However, since they are degenerate, then there is no unique basis because they all have the same eigenvalues, so linear combinations of the eigenvectors are also eigenvectors. My question is:

if we choose any arbitrary basis, will the eigenvectors be orthogonal to each other?

For example, I can choose one eigenvector to be $psi_n$ and another $alphapsi_n + betapsi_n+1$, and these are already not orthogonal.

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,0192636

asked yesterday

Josh PilipovskyJosh Pilipovsky

1011

New contributor

Josh Pilipovsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand your question, and your second sentence is not true. Choose an arbitrary basis of what? What does this basis have to do with your eigenvectors? Why isn't the last sentence of your question already an answer?
  $endgroup$
  – Qiaochu Yuan
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A matrix all of whose eigenvalues are the same number $lambda$ necessarily has the form $lambda I + N$ where $N$ is a nilpotent matrix. If it is $n times n$, then it has $n$ linearly independent eigenvectors iff $N = 0$ iff the matrix is just a scalar multiple of the identity, so any nonzero vector is an eigenvector.
  $endgroup$
  – Qiaochu Yuan
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is a "degenerate" eigenvalue?
  $endgroup$
  – David C. Ullrich
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Look, if you have all the eigenvalues the same for a matrix, then the eigenvectors are not unique. You can form linear combinations of them, and they will still be eigenvectors of the original matrix. What I'm asking is when you have already found all the eigenvectors, i) will any linear combination of them form a viable basis?, and ii) will this basis necessarily be orthonormal?
  $endgroup$
  – Josh Pilipovsky
  18 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Say you have some matrix and the eigenvalues are all the same (degenerate). Let’s say you have $N$ eigenvectors with the label $n$, call them $psi_n$. If the eigenvalues were not degenerate, then the eigenvectors would form a unique orthonormal basis for some space $mathbbV_N$. However, since they are degenerate, then there is no unique basis because they all have the same eigenvalues, so linear combinations of the eigenvectors are also eigenvectors. My question is:

if we choose any arbitrary basis, will the eigenvectors be orthogonal to each other?

For example, I can choose one eigenvector to be $psi_n$ and another $alphapsi_n + betapsi_n+1$, and these are already not orthogonal.

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,0192636

asked yesterday

Josh PilipovskyJosh Pilipovsky

1011

New contributor

Josh Pilipovsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,0192636

asked yesterday

Josh PilipovskyJosh Pilipovsky

1011

New contributor

Josh Pilipovsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,0192636

asked yesterday

Josh PilipovskyJosh Pilipovsky

1011

New contributor

Josh Pilipovsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

YuiTo Cheng

2,0192636

edited yesterday

YuiTo Cheng

2,0192636

asked yesterday

Josh PilipovskyJosh Pilipovsky

1011

New contributor

Josh Pilipovsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Josh PilipovskyJosh Pilipovsky

1011