How to interprete an analytical eigendecomposition for a polynomial... (2024)

If I configure my matrix to have a size nxn > 4x4, the symbolic math toolbox still produces analytical eigenvalues (i.e. a polynomials without the "root" expression).

Can you show us an example where this happens ? It might produce the characteristic polynomial, but how should it be possible to get analytical expressions for the roots ? It would contradict Abel-Ruffini.

However, from my experience the symbolic toolbox does not provide an analytic expression (i.e. a polynomial), if numerical methods where used and the solution is purely numerical.

I don't understand what you mean here.

So my question is how Matlab arrives at this result?

Which result ? Matlab only provides the polynomial, but not its roots.

1. The attached file "Example_matrix.mat " contains a 6x6 polynomial matrix . Where y serves as a placeholder for . Calling eig(A) produces A_roots, which are the eigenvalue-polynomials of the matrix A in terms of z and y. Atleast from my understanding Matlab therefore did in fact obtained an analytical solution for this case.
2. As discussed in: https://www.mathworks.com/matlabcentral/answers/2132636-how-does-matlabs-symbolic-math-toolbox-always-finds-an-analytical-eigendecomposition-for-arbitrary-p#answer_1478061, it seems to me that Matlab does provide a "root(...)" expression in case it can not find analytical eigenvalues.
3. From my understanding it did provide the roots, as the characteristic polynomial would be and its roots are the eigenvalues, as these are only dependent on z and y.

The attached file "Example_matrix.mat " contains a 6x6 polynomial matrix . Where y serves as a placeholder for . Calling eig(A) produces A_roots, which are the eigenvalue-polynomials of the matrix A in terms of z and y. Atleast from my understanding Matlab therefore did in fact obtained an analytical solution for this case.

eig(A) will produce a polynomial p of degree 6 by symbolically computing det(A-lambda*I) and return root(p). Is this a solution ? Formally yes, but not very helpful.

So if you consider the computation of this polynomial of degree 6 in lambda as "analytical solution", you are correct. Usually, the representation of the roots of this polynomial (that will depend on z and y) is referred to as "analytical solution" and these roots are the key of interest (because they are the eigenvalues).

Quoting my answer from below: If I use a random 6x6 matrix, the algorithm will produce a root expression - essentially telling: "I can't root this characteristic polynomial - here you have a placeholder".

But in my case: it does always return a polynomial in z and y and not a root expression. My matrix is constructed from the modelling of a physical process, and scales (depending on the inputs) to . No matter how I choose n (except n = 2), the algorithm will produce an actual polynomial. The attached example shows this - A_roots is the polynomial for the A input.

So my current understanding is that there are cases where an analytical solution for an matrix can be obtained , i.e. when exploiting the matrix structure. In this case Matlab returns an actual polynomial by computing det(A-lambda*I) and return root(p).

However, if there is no analytical solution, Matlab will just return a root-expression and essentially tell me: "I can't root this characteristic polynomial - here you have a placeholder".

Therefore, it looks like my matrix has some property, that Matlab can somehow exploit to produce the actual eigenvalues. Or why can Matlab produce a polynomial in my case, but not for a random matrix?

I cannot run your code here because it takes too long.

As you can see below for a simpler example, eig(A) returns the roots of the characteristic polynomial for A, but again in a form that doesn't help. You need to specifiy numerical values for y and z in your code and use vpa to get the numerical roots of the resulting 5th order polynomial. @John D'Errico did this in his answer.

If you really get analytical expressions for the roots of your complicated 11x11 matrix A that depend on two other symbolic variables y and z, you are really in luck and this A- matrix must be very special.

If you really get analytical expressions for the roots of your complicated 11x11 matrix A that depend on two other symbolic variables y and z, you are really in luck and this A- matrix must be very special.

Okay, that answers my question. Thanks for the effort @Torsten - I much appreciate your and @John D'Errico's help. If I find out which property that is, I will let you know.

Maybe you could insert the 11 symbolic eigenvalues you get as a graphics - I'm quite interested.

While attempting to plot the results, I have checked my own Example_matrix.mat , and the solution there is in-fact not analytical. The solution I provided did also use the "root(...)" expression to represent the eigenvalues.

Therefore, everything makes sense now and fits into the in-depth explanation that @John D'Errico provided earlier.

I have re-run a few tests with my framework to check if I can find any case for > 4 matrix with a polynomial solution. However, all of them ended up with "root(...)" solutions.

Therefore: You were correct all along, I just misread the results for the 6x6 case. Again thanks for your patience in explaining the matter.

syms x

solve(x^6 - 1 == 0)

load Example_matrix

whos

A

syms y z

vpa(eig(subs(A,[y,z],[3,17])))

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