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Unable to evaluate Eigenvalues and Eigenvectors for a matrix (2)

Unable to evaluate Eigenvalues and Eigenvectors for a matrixProblem with Eigenvectors when given a matrix containing approximate numbers and symbolsIs Eigensystem::eivin message a bug?Seemingly wrong eigenvectors for numerical matrix whose elements differ in scale by orders of magnitudeInverse of a 3x3 MatrixConvert, using the Pauli matrices, an $n times m$ matrix of quaternions into a $2 n times 2 m$ matrix with complex entries, and vice versaEigenvalues and eigenvectors of tensorsNo eigenvectors coming for a very simple* matrixUnable to evaluate Eigenvalues and Eigenvectors for a matrixBasis for unstable manifold of a matrixDensity map for complex and imaginary parts of eigenvalues on one graph

2

$begingroup$

I have posted a similar question last year pertaining to this issue. Here's a link to my post together with the solution given: Unable to evaluate Eigenvalues and Eigenvectors for a matrix

I have tried the methods in my previous posts but to no avail. Here's the problem: I have the following 3x3 matrix

m = -γ/2, -I*g1, -I*Exp[-I*α]*g3, -I*g1, -(κ1)/2, -I*g2, -I*Exp[I*α]*g3, -I*g2, -(κ2)/2]

where I represents the complex identity Sqrt[-1]. I wish to find the eigenvectors for the matrix for two different alpha values. For α = π/2, simply doing (after manually replacing α with π/2)

Eigenvectors[m, Cubics->True]

Returns the appropriate (albeit long) eigenvectors. Now however, if I change my α to α = π and run

 Eigenvectors[m, Cubics->True]

I am returned with

...Eigenvectors: Unable to find all eigenvectors

Which is the similar issue encountered in the link that I provided above a while ago. I proceed to perform the same fix detailed in that question. Namely

Simplify[Eigenvectors[mchiral /. Complex[0, -1] -> mi, Cubics -> True] /. mi -> -I];

and I am still returned with the same error. Namely

...Eigenvectors: Unable to find all eigenvectors

What is the problem here?

matrix eigenvalues

share|improve this question

edited 4 hours ago

corey979

20.9k64282

asked 4 hours ago

kowalskikowalski

1559

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is no problem: imgur.com/a/ALdYCou Except that you have some brackets misplaced in the definition of m that I fixed – but check if the form is the desired one.
  $endgroup$
  – corey979
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @corey979 Interesting. I am on version 11.3 for macOS and I do get the error message (even after fixing the brackets).
  $endgroup$
  – Henrik Schumacher
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How many times have I advocated for providing the $Version one is using... I'm on 10.4 and there is no problem, as showed. Indeed, there is an error in 11.3. No idea what version the OP is using. Unless he clarifies there is virtually no problem.
  $endgroup$
  – corey979
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @corey979 Apologies but I'm using version 11.3. The error still persists even after the bracket fix. I don't know what the problem is
  $endgroup$
  – kowalski
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I have posted a similar question last year pertaining to this issue. Here's a link to my post together with the solution given: Unable to evaluate Eigenvalues and Eigenvectors for a matrix

I have tried the methods in my previous posts but to no avail. Here's the problem: I have the following 3x3 matrix

m = -γ/2, -I*g1, -I*Exp[-I*α]*g3, -I*g1, -(κ1)/2, -I*g2, -I*Exp[I*α]*g3, -I*g2, -(κ2)/2]

where I represents the complex identity Sqrt[-1]. I wish to find the eigenvectors for the matrix for two different alpha values. For α = π/2, simply doing (after manually replacing α with π/2)

Eigenvectors[m, Cubics->True]

Returns the appropriate (albeit long) eigenvectors. Now however, if I change my α to α = π and run

 Eigenvectors[m, Cubics->True]

I am returned with

...Eigenvectors: Unable to find all eigenvectors

Which is the similar issue encountered in the link that I provided above a while ago. I proceed to perform the same fix detailed in that question. Namely

Simplify[Eigenvectors[mchiral /. Complex[0, -1] -> mi, Cubics -> True] /. mi -> -I];

and I am still returned with the same error. Namely

...Eigenvectors: Unable to find all eigenvectors

What is the problem here?

matrix eigenvalues

share|improve this question

edited 4 hours ago

corey979

20.9k64282

asked 4 hours ago

kowalskikowalski

1559

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is no problem: imgur.com/a/ALdYCou Except that you have some brackets misplaced in the definition of m that I fixed – but check if the form is the desired one.
  $endgroup$
  – corey979
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @corey979 Interesting. I am on version 11.3 for macOS and I do get the error message (even after fixing the brackets).
  $endgroup$
  – Henrik Schumacher
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How many times have I advocated for providing the $Version one is using... I'm on 10.4 and there is no problem, as showed. Indeed, there is an error in 11.3. No idea what version the OP is using. Unless he clarifies there is virtually no problem.
  $endgroup$
  – corey979
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @corey979 Apologies but I'm using version 11.3. The error still persists even after the bracket fix. I don't know what the problem is
  $endgroup$
  – kowalski
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I have posted a similar question last year pertaining to this issue. Here's a link to my post together with the solution given: Unable to evaluate Eigenvalues and Eigenvectors for a matrix

I have tried the methods in my previous posts but to no avail. Here's the problem: I have the following 3x3 matrix

m = -γ/2, -I*g1, -I*Exp[-I*α]*g3, -I*g1, -(κ1)/2, -I*g2, -I*Exp[I*α]*g3, -I*g2, -(κ2)/2]

where I represents the complex identity Sqrt[-1]. I wish to find the eigenvectors for the matrix for two different alpha values. For α = π/2, simply doing (after manually replacing α with π/2)

Eigenvectors[m, Cubics->True]

Returns the appropriate (albeit long) eigenvectors. Now however, if I change my α to α = π and run

 Eigenvectors[m, Cubics->True]

I am returned with

...Eigenvectors: Unable to find all eigenvectors

Which is the similar issue encountered in the link that I provided above a while ago. I proceed to perform the same fix detailed in that question. Namely

Simplify[Eigenvectors[mchiral /. Complex[0, -1] -> mi, Cubics -> True] /. mi -> -I];

and I am still returned with the same error. Namely

...Eigenvectors: Unable to find all eigenvectors

What is the problem here?

matrix eigenvalues

share|improve this question

edited 4 hours ago

corey979

20.9k64282

asked 4 hours ago

kowalskikowalski

1559

$endgroup$

m = -γ/2, -I*g1, -I*Exp[-I*α]*g3, -I*g1, -(κ1)/2, -I*g2, -I*Exp[I*α]*g3, -I*g2, -(κ2)/2]

Eigenvectors[m, Cubics->True]

 Eigenvectors[m, Cubics->True]

...Eigenvectors: Unable to find all eigenvectors

Simplify[Eigenvectors[mchiral /. Complex[0, -1] -> mi, Cubics -> True] /. mi -> -I];

...Eigenvectors: Unable to find all eigenvectors

share|improve this question

edited 4 hours ago

corey979

20.9k64282

asked 4 hours ago

kowalskikowalski

1559

share|improve this question

edited 4 hours ago

corey979

20.9k64282

asked 4 hours ago

kowalskikowalski

1559

edited 4 hours ago

corey979

20.9k64282

edited 4 hours ago

corey979

20.9k64282

asked 4 hours ago

kowalskikowalski

1559

asked 4 hours ago

kowalskikowalski

1559

1 Answer
1

active

oldest

votes

2

$begingroup$

I have o clue why this did not work. However, this old-fashioned method seems to work

m = 
 -γ/2, -I g1, -I Exp[-I α] g3, 
 -I g1, -(κ1)/2, -I g2, 
 -I Exp[I α] g3, -I g2, -(κ2)/2
 ;

a = m /. α -> π;
λ = Eigenvalues[a] // ToRadicals;
U = Flatten[NullSpace[a - # IdentityMatrix[3]] & /@ λ, 1];

Simplify[a.Transpose[U] - Transpose[U].DiagonalMatrix[λ]]

0, 0, 0, 0, 0, 0, 0, 0, 0

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Henrik SchumacherHenrik Schumacher

56.7k577157

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The eigenvalues seems to be fine. It's long but it prints. The eigenvectors on the other hand, are still unobtainable. I'm starting to think if this is alpha dependent since it works for pi/2 but not pi. But that's just silly and shouldn't happen
  $endgroup$
  – kowalski
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kowalski U produced by the code above contains the eigenvectors.
  $endgroup$
  – Henrik Schumacher
  3 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

I have o clue why this did not work. However, this old-fashioned method seems to work

m = 
 -γ/2, -I g1, -I Exp[-I α] g3, 
 -I g1, -(κ1)/2, -I g2, 
 -I Exp[I α] g3, -I g2, -(κ2)/2
 ;

a = m /. α -> π;
λ = Eigenvalues[a] // ToRadicals;
U = Flatten[NullSpace[a - # IdentityMatrix[3]] & /@ λ, 1];

Simplify[a.Transpose[U] - Transpose[U].DiagonalMatrix[λ]]

0, 0, 0, 0, 0, 0, 0, 0, 0

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Henrik SchumacherHenrik Schumacher

56.7k577157

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The eigenvalues seems to be fine. It's long but it prints. The eigenvectors on the other hand, are still unobtainable. I'm starting to think if this is alpha dependent since it works for pi/2 but not pi. But that's just silly and shouldn't happen
  $endgroup$
  – kowalski
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kowalski U produced by the code above contains the eigenvectors.
  $endgroup$
  – Henrik Schumacher
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I have o clue why this did not work. However, this old-fashioned method seems to work

m = 
 -γ/2, -I g1, -I Exp[-I α] g3, 
 -I g1, -(κ1)/2, -I g2, 
 -I Exp[I α] g3, -I g2, -(κ2)/2
 ;

a = m /. α -> π;
λ = Eigenvalues[a] // ToRadicals;
U = Flatten[NullSpace[a - # IdentityMatrix[3]] & /@ λ, 1];

Simplify[a.Transpose[U] - Transpose[U].DiagonalMatrix[λ]]

0, 0, 0, 0, 0, 0, 0, 0, 0

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Henrik SchumacherHenrik Schumacher

56.7k577157

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The eigenvalues seems to be fine. It's long but it prints. The eigenvectors on the other hand, are still unobtainable. I'm starting to think if this is alpha dependent since it works for pi/2 but not pi. But that's just silly and shouldn't happen
  $endgroup$
  – kowalski
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kowalski U produced by the code above contains the eigenvectors.
  $endgroup$
  – Henrik Schumacher
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I have o clue why this did not work. However, this old-fashioned method seems to work

m = 
 -γ/2, -I g1, -I Exp[-I α] g3, 
 -I g1, -(κ1)/2, -I g2, 
 -I Exp[I α] g3, -I g2, -(κ2)/2
 ;

a = m /. α -> π;
λ = Eigenvalues[a] // ToRadicals;
U = Flatten[NullSpace[a - # IdentityMatrix[3]] & /@ λ, 1];

Simplify[a.Transpose[U] - Transpose[U].DiagonalMatrix[λ]]

0, 0, 0, 0, 0, 0, 0, 0, 0

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Henrik SchumacherHenrik Schumacher

56.7k577157

$endgroup$

m = 
 -γ/2, -I g1, -I Exp[-I α] g3, 
 -I g1, -(κ1)/2, -I g2, 
 -I Exp[I α] g3, -I g2, -(κ2)/2
 ;

a = m /. α -> π;
λ = Eigenvalues[a] // ToRadicals;
U = Flatten[NullSpace[a - # IdentityMatrix[3]] & /@ λ, 1];

Simplify[a.Transpose[U] - Transpose[U].DiagonalMatrix[λ]]

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Henrik SchumacherHenrik Schumacher

56.7k577157

answered 4 hours ago

Henrik SchumacherHenrik Schumacher

56.7k577157

answered 4 hours ago

Henrik SchumacherHenrik Schumacher

56.7k577157