Calculating eigenvalues of one matrix

AI Thread Summary
The discussion revolves around calculating the eigenvalues of a complex 4x4 matrix using MATLAB and Mathematica, which initially struggle with the complexity of the matrix. Users suggest that the eigenvalues can be derived from a quartic equation, and applying the ToRadicals function in Mathematica can yield explicit solutions. After successfully obtaining the eigenvalues, the user seeks assistance in expanding these values for small variables x, y, and z. The recommended approach for small argument expansion is to use the Series function in Mathematica, which allows for the expansion of the eigenvalues in terms of the small variables. The conversation highlights the challenges and solutions in handling complex matrix computations.
quin
Messages
50
Reaction score
0
hi friends

i want to find eigenvalues of a 4*4 matrix but niether MATLAB nor MATHEMATICA can't solve it. Both of programs said that eiganvalues of matrix is too complicated and have infinite sentences.
now what can i do?is there anyway that simplify the steps for MATLAB or mathematica?

thanks for your help

the matrix is here:
 

Attachments

Physics news on Phys.org
Mathematica gives four eigenvalues in a few seconds for

a = {{0, 1 + E^(-I*x) + E^(-I*z) + E^(-I*(x + z)), 1 + E^(-I*x) + E^(-I*y) + E^(-I*(x + y)), 1 + E^(-I*y) + E^(-I*z) + E^(-I*(y + z))},
{1 + E^(I*x) + E^(I*z) + E^(I*(x + z)), 0, 1 + E^(-I*y) + E^(I*z) + E^(-I*(y - z)), 1 + E^(I*x) + E^(-I*y) + E^(I*(x - y))},
{1 + E^(I*x) + E^(I*y) + E^(I*(x + y)), 1 + E^(I*y) + E^(-I*z) + E^(I*(y - z)), 0, 1 + E^(I*x) + E^(-I*z) + E^(I*(x - z))},
{1 + E^(I*y) + E^(I*z) + E^(I*(y + z)), 1 + E^(-I*x) + E^(I*y) + E^(-I*(x - y)), 1 + E^(-I*x) + E^(I*z) +E^(-I*(x - z)), 0}};
Eigenvalues[a]

Each of those is a root of a quartic equation. Using ToRadicals on each of those will give you the explicit solution
 
Bill Simpson said:
Mathematica gives four eigenvalues in a few seconds for

a = {{0, 1 + E^(-I*x) + E^(-I*z) + E^(-I*(x + z)), 1 + E^(-I*x) + E^(-I*y) + E^(-I*(x + y)), 1 + E^(-I*y) + E^(-I*z) + E^(-I*(y + z))},
{1 + E^(I*x) + E^(I*z) + E^(I*(x + z)), 0, 1 + E^(-I*y) + E^(I*z) + E^(-I*(y - z)), 1 + E^(I*x) + E^(-I*y) + E^(I*(x - y))},
{1 + E^(I*x) + E^(I*y) + E^(I*(x + y)), 1 + E^(I*y) + E^(-I*z) + E^(I*(y - z)), 0, 1 + E^(I*x) + E^(-I*z) + E^(I*(x - z))},
{1 + E^(I*y) + E^(I*z) + E^(I*(y + z)), 1 + E^(-I*x) + E^(I*y) + E^(-I*(x - y)), 1 + E^(-I*x) + E^(I*z) +E^(-I*(x - z)), 0}};
Eigenvalues[a]

Each of those is a root of a quartic equation. Using ToRadicals on each of those will give you the explicit solution

thank you so much dear it worked

But I have another question too
I found 4 eigenvalues and they are 4 sentences in terms of x,y,z

now I want to expand all of them (4 of them) for "small x and small y and small z"

can you give me the the suitable formula for mathematica for small argument expansion?

thank you
 
Last edited:
If you had an "ordinary" expression then

Limit[Limit[Limit[p, x -> 0], y -> 0], z -> 0]

would give you the limit as your three variables go to zero, but Mathematica ToRules and Limit doesn't seem happy that everything you have is in terms of of complex exponentials.

Perhaps this will help you

Eigenvalues[a] //. {x -> 10^-9, y -> 10^-9, z -> 10^-9}

returns

{-2 - E^(-I/1000000000) - E^(I/1000000000),
-2 - E^(-I/1000000000) - E^(I/1000000000),
-2 - E^(-I/1000000000) - E^(I/1000000000),
3*(2 + E^(-I/1000000000) + E^(I/1000000000))}

BUT that is only approaching zero from one special direction and that direction is going to result in a lot of cancellations in your eigenvalues.

Perhaps you need to think carefully about exactly what you are trying to accomplish. That might give you an idea of what direction you want to approach this from. No pun intended.
 
Last edited:
Thanks for your answer
and also now I found the formula "series" that expands the expression in terms of small x and small y and small z :
Series[..., {x,0,2},{y,0,2}, {z,0,2}]
 

Similar threads

Replies
12
Views
6K
Replies
1
Views
3K
Replies
12
Views
2K
Replies
0
Views
1K
Replies
10
Views
3K
Replies
4
Views
3K
Replies
5
Views
1K
Back
Top