Eigenvalues and eigenvectors of a matrix gives z1 in the answer... what does it means ?

Hello all, hope you guys are doing good.

I would like to find the eigenvalues of a matrix, but the answer looks confusing.

Here is the code:

syms w
MK = [0.02*w^2-2000 1000 0 0 0 0 0 0 0 0
    1000 0.02*w^2-2000 1000 0 0 0 0 0 0 0
    0 1000 0.02*w^2-2000 1000 0 0 0 0 0 0
    0 0 1000 0.02*w^2-2000 1000 0 0 0 0 0
    0 0 0 1000 0.02*w^2-2000 1000 0 0 0 0
    0 0 0 0 1000 0.02*w^2-2000 1000 0 0 0
    0 0 0 0 0 1000 0.02*w^2-2000 1000 0 0
    0 0 0 0 0 0 1000 0.02*w^2-2000 1000 0
    0 0 0 0 0 0 0 1000 0.02*w^2-2000 1000
    0 0 0 0 0 0 0 0 1000 0.02*w^2-2000]
E = eig(MK)

This problem has 8 masses so in it is supposed to give me 8 eigenvalues

I did the previous question ( the number of masses were 4 ):

This is the matrix of the previous question:

and the first eigenvalue of the previous question is:

but the answer for the matrix with 8 masses looks like weird:

E =
 
     root(z1^5 - z1^4*(w^2/10 - 9000) + z1^3*(- 720*w^2 + w^4/250 + 28000000) - z1^2*(1680000*w^2 - (108*w^4)/5 + w^6/12500 - 35000000000) + z1*(- 1400000000*w^2 + 33600*w^4 - (36*w^6)/125 + w^8/1250000 + 15000000000000) - w^10/312500000 + 14000000*w^4 + (9*w^8)/6250 - 224*w^6 - 300000000000*w^2 + 1000000000000000, z1, 1)
     root(z1^5 - z1^4*(w^2/10 - 9000) + z1^3*(- 720*w^2 + w^4/250 + 28000000) - z1^2*(1680000*w^2 - (108*w^4)/5 + w^6/12500 - 35000000000) + z1*(- 1400000000*w^2 + 33600*w^4 - (36*w^6)/125 + w^8/1250000 + 15000000000000) - w^10/312500000 + 14000000*w^4 + (9*w^8)/6250 - 224*w^6 - 300000000000*w^2 + 1000000000000000, z1, 2)
     root(z1^5 - z1^4*(w^2/10 - 9000) + z1^3*(- 720*w^2 + w^4/250 + 28000000) - z1^2*(1680000*w^2 - (108*w^4)/5 + w^6/12500 - 35000000000) + z1*(- 1400000000*w^2 + 33600*w^4 - (36*w^6)/125 + w^8/1250000 + 15000000000000) - w^10/312500000 + 14000000*w^4 + (9*w^8)/6250 - 224*w^6 - 300000000000*w^2 + 1000000000000000, z1, 3)
     root(z1^5 - z1^4*(w^2/10 - 9000) + z1^3*(- 720*w^2 + w^4/250 + 28000000) - z1^2*(1680000*w^2 - (108*w^4)/5 + w^6/12500 - 35000000000) + z1*(- 1400000000*w^2 + 33600*w^4 - (36*w^6)/125 + w^8/1250000 + 15000000000000) - w^10/312500000 + 14000000*w^4 + (9*w^8)/6250 - 224*w^6 - 300000000000*w^2 + 1000000000000000, z1, 4)
     root(z1^5 - z1^4*(w^2/10 - 9000) + z1^3*(- 720*w^2 + w^4/250 + 28000000) - z1^2*(1680000*w^2 - (108*w^4)/5 + w^6/12500 - 35000000000) + z1*(- 1400000000*w^2 + 33600*w^4 - (36*w^6)/125 + w^8/1250000 + 15000000000000) - w^10/312500000 + 14000000*w^4 + (9*w^8)/6250 - 224*w^6 - 300000000000*w^2 + 1000000000000000, z1, 5)
 root(z1^5 - z1^4*(w^2/10 - 11000) + z1^3*(- 880*w^2 + w^4/250 + 44000000) - z1^2*(2640000*w^2 - (132*w^4)/5 + w^6/12500 - 77000000000) + z1*(- 3080000000*w^2 + 52800*w^4 - (44*w^6)/125 + w^8/1250000 + 55000000000000) - 1100000000000*w^2 - w^10/312500000 + 30800000*w^4 + (11*w^8)/6250 - 352*w^6 + 11000000000000000, z1, 1)
 root(z1^5 - z1^4*(w^2/10 - 11000) + z1^3*(- 880*w^2 + w^4/250 + 44000000) - z1^2*(2640000*w^2 - (132*w^4)/5 + w^6/12500 - 77000000000) + z1*(- 3080000000*w^2 + 52800*w^4 - (44*w^6)/125 + w^8/1250000 + 55000000000000) - 1100000000000*w^2 - w^10/312500000 + 30800000*w^4 + (11*w^8)/6250 - 352*w^6 + 11000000000000000, z1, 2)
 root(z1^5 - z1^4*(w^2/10 - 11000) + z1^3*(- 880*w^2 + w^4/250 + 44000000) - z1^2*(2640000*w^2 - (132*w^4)/5 + w^6/12500 - 77000000000) + z1*(- 3080000000*w^2 + 52800*w^4 - (44*w^6)/125 + w^8/1250000 + 55000000000000) - 1100000000000*w^2 - w^10/312500000 + 30800000*w^4 + (11*w^8)/6250 - 352*w^6 + 11000000000000000, z1, 3)
 root(z1^5 - z1^4*(w^2/10 - 11000) + z1^3*(- 880*w^2 + w^4/250 + 44000000) - z1^2*(2640000*w^2 - (132*w^4)/5 + w^6/12500 - 77000000000) + z1*(- 3080000000*w^2 + 52800*w^4 - (44*w^6)/125 + w^8/1250000 + 55000000000000) - 1100000000000*w^2 - w^10/312500000 + 30800000*w^4 + (11*w^8)/6250 - 352*w^6 + 11000000000000000, z1, 4)
 root(z1^5 - z1^4*(w^2/10 - 11000) + z1^3*(- 880*w^2 + w^4/250 + 44000000) - z1^2*(2640000*w^2 - (132*w^4)/5 + w^6/12500 - 77000000000) + z1*(- 3080000000*w^2 + 52800*w^4 - (44*w^6)/125 + w^8/1250000 + 55000000000000) - 1100000000000*w^2 - w^10/312500000 + 30800000*w^4 + (11*w^8)/6250 - 352*w^6 + 11000000000000000, z1, 5)

What z1 means ? my professor told me to solve this analytically but I have no idea how to obtain the value of w (omega).

Help Please !

Thank you !

[Christine Tobler]

The expression

root(z1^5 - z1^4*(w^2/10 - 11000) +... + 11000000000000000, z1, 5)

means that this eigenvalue is the 5th root (solution) of the polynom given as the first input in z1. See root doc page.

Polynomials with a degree greater than 4 do not have explicit solutions, so this is the most direct form that the eigenvalues can be represented in symbolically.

[Rodrigo Pena]

So basically there is no much I can do right ?

If I want to solve this numerically, how can I proceed ?

Thanks !

[Christine Tobler]

If you insert a specific value for w, you can solve the polynomial numerically:

>> vpa(subs(E, w, 4))
 ans =
 -80.694052771005220219263885867345
-689.95853210942987188614985506741
-1715.0503234534297191124146627673
-2830.5100260037728510585482984592
-3682.1870656623623377236232978387
-317.17293433763766227637670216126
-1168.8499739962271489414517015408
-2284.3096765465702808875853372327
-3309.4014678905701281138501449326
-3918.6659472289947797807361141327

Although at that point, you might as well substitute w in the original matrix and solve that with the numerical EIG method:

>> eig(subs(MK, w, 4))
 ans =
 -80.694052771005220219263885867345
-689.95853210942987188614985506741
-1715.0503234534297191124146627673
-2830.5100260037728510585482984592
-3682.1870656623623377236232978387
-317.17293433763766227637670216126
-1168.8499739962271489414517015408
-2284.3096765465702808875853372327
-3309.4014678905701281138501449326
-3918.6659472289947797807361141327

[Rodrigo Pena]

Ok, In this case you assumed that w = 4 ?

[Bjorn Gustavsson]

You might get some "half-analytical" results from the tools in chebfun - it is an enormously brilliant tool, it has a lot of clever functionalities, and some examples of eigenvalue-problems - that might be similar enough to your problem.

[Steven Lord]

Yes. To be a bit pedantic Christine did not assume that w was equal to 4, she told MATLAB that it should substitute the value 4 for w in the symbolic variables E and MK. That's the purpose of the subs function.

[Rodrigo Pena]

I see...

My professor just told me that I should use this command here:

[V,D] = eig(A,B) returns diagonal matrix Dof generalized eigenvalues and full matrix V whose columns are the corresponding right eigenvectors, so that A*V = B*V*D.

But, what would be the matrix A and B ?

This screeshot is from the previous question... the actual question instead of 4 masses is 10

[Christine Tobler]

I don't know the context, but I don't think your professor meant for you to use the symbolic toolbox with omega as an unknown.

Think about if you can rearrange the last equation you have here to fit the A*v = lambda*B*v formula for generalized eig, with no unknowns in matrices A and B.

[Rodrigo Pena]

I see... he told me that the eigenvalues are the values of omega squared.

I used the syntax E=eig(MK) for a matrix 15x15 (same thing as before but now the number of masses are 15. And the results that I got was:

Which are consistent because it does not show that weird z1. and it is a matrix 15x15 so teorically the polynomial degree would be higher than the one with 10x10 right ?

This is nuts. I don't know what to do.

[Christine Tobler]

Don't think about the implementation or the z1 right now, using symbolic at all was the wrong approach.

Just take pen and paper and your equation (the one after "Rearranging:"), and try to reformulate this so you it looks like A*x = lambda*B*x, where A and B don't contain omega (the omega should be represented by lambda here, which is the eigenvalue).

Once you have that, it should be clear how you can solve using EIG.

[Rodrigo Pena]

Like this?

To obtain the eigenvectors do I have to use the symtax with the same matrix form ? (A*x = λ*B*x)

[Christine Tobler]

Yes, that's it. So you have two matrices A and B now, which you can pass to EIG. The output will give you the eigenvalues lambda (from which you can then compute omega, since lambda = omega^2 here). And you can get matching A1, A2, A3, A4 from the eigenvectors for that lambda.

[Rodrigo Pena]

Ok, Do I have to say that my matrix A is negative to Matlab ?

[Rodrigo Pena]

Or maybe just change signs of each number in the matrix ?

[Christine Tobler]

You can do either way, put minus in front of the matrix or change signs. You can even move it over to the omega, say lambda = -omega^2 instead of lambda = omega^2, both work just as well.

[Rodrigo Pena]

Perfect, Thank you so much !!