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Is it true that all symmetric matrices have eigenbasis?

Symmetric MatricesSymmetric matrix is always diagonalizable?How to find a mutually orthogonal basis for symmetric matrices where 2 of the eigenvectors are equalFinding orthonormal eigenvectors of symmetric matricesComplex symmetric matrix orthogonal eigenvectorsProving symmetric matrices are diagonalizable using fact eigenvectors must be orthogonalwhy symmetric matrix is always diagonalizable even when it has repeated eigenvalues?Do Real Symmetric Matrices have 'n' linearly independent eigenvectors?Eigenvalues, orthonormal eigenvectorsProof for why symmetric matrices are only orthogonally diagonalizable

0

$begingroup$

I know that a symmetric matrix has a set of orthogonal eigenvectors if all its eigenvalues are distinct. As far as I know, their eigenvalues might not be distinct. However, some sources claim that symmetric matrices always have eigenbasis.

I know that we can get orthonormal basis using Gram Schmidt on the eigenvectors of each basis. However, I don't think that it guarantees the resultant vectors are still eigenvectors. Shouldn't the Gram Schmidt process turns the eigenvectors into non eigenvectors? If not how to prove that the resultant vectors are still eigenvectors?

So, my question then is if symmetric matrices always have eigenbasis regardless of whether its eigenvalues are distinct? Best if a calculus-free proof for whether or not symmetric matrices have eigenbasis can be given.

linear-algebra matrices eigenvalues-eigenvectors orthogonality symmetric-matrices

share|cite|improve this question

edited Nov 17 '17 at 5:45

user10024395

asked Nov 16 '17 at 17:29

user10024395user10024395

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is true over reals. The proof is simple: find some eigenvector (it exists) v and continue by induction on $v^perp$ (which is $A$-invariant by the symmetry of $A$)
  $endgroup$
  – Peter Franek
  Nov 16 '17 at 17:47
  

  
  
  
  

add a comment | 

0

$begingroup$

I know that a symmetric matrix has a set of orthogonal eigenvectors if all its eigenvalues are distinct. As far as I know, their eigenvalues might not be distinct. However, some sources claim that symmetric matrices always have eigenbasis.

I know that we can get orthonormal basis using Gram Schmidt on the eigenvectors of each basis. However, I don't think that it guarantees the resultant vectors are still eigenvectors. Shouldn't the Gram Schmidt process turns the eigenvectors into non eigenvectors? If not how to prove that the resultant vectors are still eigenvectors?

So, my question then is if symmetric matrices always have eigenbasis regardless of whether its eigenvalues are distinct? Best if a calculus-free proof for whether or not symmetric matrices have eigenbasis can be given.

linear-algebra matrices eigenvalues-eigenvectors orthogonality symmetric-matrices

share|cite|improve this question

edited Nov 17 '17 at 5:45

user10024395

asked Nov 16 '17 at 17:29

user10024395user10024395

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is true over reals. The proof is simple: find some eigenvector (it exists) v and continue by induction on $v^perp$ (which is $A$-invariant by the symmetry of $A$)
  $endgroup$
  – Peter Franek
  Nov 16 '17 at 17:47
  

  
  
  
  

add a comment | 

$begingroup$

I know that a symmetric matrix has a set of orthogonal eigenvectors if all its eigenvalues are distinct. As far as I know, their eigenvalues might not be distinct. However, some sources claim that symmetric matrices always have eigenbasis.

I know that we can get orthonormal basis using Gram Schmidt on the eigenvectors of each basis. However, I don't think that it guarantees the resultant vectors are still eigenvectors. Shouldn't the Gram Schmidt process turns the eigenvectors into non eigenvectors? If not how to prove that the resultant vectors are still eigenvectors?

So, my question then is if symmetric matrices always have eigenbasis regardless of whether its eigenvalues are distinct? Best if a calculus-free proof for whether or not symmetric matrices have eigenbasis can be given.

linear-algebra matrices eigenvalues-eigenvectors orthogonality symmetric-matrices

share|cite|improve this question

edited Nov 17 '17 at 5:45

user10024395

asked Nov 16 '17 at 17:29

user10024395user10024395

1

$endgroup$

share|cite|improve this question

edited Nov 17 '17 at 5:45

user10024395

asked Nov 16 '17 at 17:29

user10024395user10024395

1

share|cite|improve this question

edited Nov 17 '17 at 5:45

user10024395

asked Nov 16 '17 at 17:29

user10024395user10024395

1

asked Nov 16 '17 at 17:29

user10024395user10024395

1

asked Nov 16 '17 at 17:29

user10024395user10024395

1

2 Answers
2

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oldest

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0

$begingroup$

Yes, each real symmetric matrices is orthogonaly equivalent to a diagonal matrix. In other words, you can find an orthonormal basis of eigenvectors. This is the spectral theorem for symmetric matrices.

share|cite|improve this answer

answered Nov 16 '17 at 17:35

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

add a comment | 

0

$begingroup$

Yes, a symmetric matrix always has an eigenbasis: as an $n times n$ symmetric matrix always has $n$ eigenvectors (spectral theorem), which can be made orthogonal by the Gram-Schmidt theorem. This proves that the eigenbasis doesn't depend on repeated or distinct eigenvalues.

share|cite|improve this answer

edited Mar 10 at 14:16

Glorfindel

3,42981830

answered Mar 10 at 13:59

Kanika RajainKanika Rajain

246

$endgroup$

add a comment | 

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0

$begingroup$

Yes, each real symmetric matrices is orthogonaly equivalent to a diagonal matrix. In other words, you can find an orthonormal basis of eigenvectors. This is the spectral theorem for symmetric matrices.

share|cite|improve this answer

answered Nov 16 '17 at 17:35

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

add a comment | 

0

$begingroup$

Yes, each real symmetric matrices is orthogonaly equivalent to a diagonal matrix. In other words, you can find an orthonormal basis of eigenvectors. This is the spectral theorem for symmetric matrices.

share|cite|improve this answer

answered Nov 16 '17 at 17:35

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

add a comment | 

$begingroup$

Yes, each real symmetric matrices is orthogonaly equivalent to a diagonal matrix. In other words, you can find an orthonormal basis of eigenvectors. This is the spectral theorem for symmetric matrices.

share|cite|improve this answer

answered Nov 16 '17 at 17:35

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

share|cite|improve this answer

answered Nov 16 '17 at 17:35

José Carlos SantosJosé Carlos Santos

167k22132235

answered Nov 16 '17 at 17:35

José Carlos SantosJosé Carlos Santos

167k22132235

answered Nov 16 '17 at 17:35

José Carlos SantosJosé Carlos Santos

167k22132235

0

$begingroup$

Yes, a symmetric matrix always has an eigenbasis: as an $n times n$ symmetric matrix always has $n$ eigenvectors (spectral theorem), which can be made orthogonal by the Gram-Schmidt theorem. This proves that the eigenbasis doesn't depend on repeated or distinct eigenvalues.

share|cite|improve this answer

edited Mar 10 at 14:16

Glorfindel

3,42981830

answered Mar 10 at 13:59

Kanika RajainKanika Rajain

246

$endgroup$

add a comment | 

0

$begingroup$

Yes, a symmetric matrix always has an eigenbasis: as an $n times n$ symmetric matrix always has $n$ eigenvectors (spectral theorem), which can be made orthogonal by the Gram-Schmidt theorem. This proves that the eigenbasis doesn't depend on repeated or distinct eigenvalues.

share|cite|improve this answer

edited Mar 10 at 14:16

Glorfindel

3,42981830

answered Mar 10 at 13:59

Kanika RajainKanika Rajain

246

$endgroup$

add a comment | 

$begingroup$

Yes, a symmetric matrix always has an eigenbasis: as an $n times n$ symmetric matrix always has $n$ eigenvectors (spectral theorem), which can be made orthogonal by the Gram-Schmidt theorem. This proves that the eigenbasis doesn't depend on repeated or distinct eigenvalues.

share|cite|improve this answer

edited Mar 10 at 14:16

Glorfindel

3,42981830

answered Mar 10 at 13:59

Kanika RajainKanika Rajain

246

$endgroup$

share|cite|improve this answer

edited Mar 10 at 14:16

Glorfindel

3,42981830

answered Mar 10 at 13:59

Kanika RajainKanika Rajain

246

edited Mar 10 at 14:16

Glorfindel

3,42981830

edited Mar 10 at 14:16

Glorfindel

3,42981830

answered Mar 10 at 13:59

Kanika RajainKanika Rajain

246

answered Mar 10 at 13:59

Kanika RajainKanika Rajain

246