# How to interprete an analytical eigendecomposition for a polynomial matrix with size nxn > 4x4 using the Symbolic Math Toolbox.

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Bastian Loß 2024 年 6 月 27 日

Hi,
I am using the Symbolic Math Toolbox to calculate the analytical eigenvalues of a polynomial matrix A(z), that is generated in my physical use-case. Depending on the input parameters, the matrix can scale in size from 1x1 to any nxn.
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). How does Matlab calculate the analytical Eigenvalues in this particular case?
According to an answer to my previous question: https://www.mathworks.com/matlabcentral/answers/2132636-how-does-matlabs-symbolic-math-toolbox-always-finds-an-analytical-eigendecomposition-for-arbitrary-p#answer_1478061, Matlab uses numerical methods to find the roots of the characteristic polynomial.
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. So my question is how Matlab arrives at this result?
1. Does Matlab indeed use a numerical approximation?
2. Does Matlab do some "tricks" in the background to reformulate my matrix into some analytically solvable matrix?
I also have attached a copy of my Matrix in a 6x6 configuration and the corresponding result.
Note: the variable "y" in my matrix represents 1/z, but this shouldn't make a difference.
My code does the following:
syms z y
A = A;
A_roots = eig(A);
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Torsten 2024 年 6 月 28 日
Maybe you could insert the 11 symbolic eigenvalues you get as a graphics - I'm quite interested.
Bastian Loß 2024 年 6 月 28 日

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.

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### 採用された回答

John D'Errico 2024 年 6 月 27 日

Ok, do you understand that you CANNOT compute the roots of a polynomial, if the degree of that polynomial is greater than 4? Well, at least, unless the roots are trivial. For example, yes, this works.
syms x
solve(x^6 - 1 == 0)
ans =
But in general, if the polynomial is non-trivial, then there is no general solution, and this is provably true? (Talk to some folks named Abel and Ruffini if that upsets you. They are both long dead of course, so you may need a Ouija board, or something as useful.)
Next, you need to understand that computing the eigenvalues of a matrix is mathematically equivalent to solving for the roots of a polynomial. So a 5x5 or 6x6 matrix has a chacteristic polynomial, and the roots of that polynomial are the same as the eigenvalues of the matrix. There is a complete correspondence between the two problems. If you can solve one, you can solve the other. But I just got done telling you that you CANNOT compute the roots of a general symbolic polynomial of degree 5 or higher, where the result has an algebraic solution. For degree 4 or below, the problem is solved, and the solutions have also existed for many years. That means, effectively, the eigenvalues of even a fully general 4x4 matrix can be found in algebraic form. They will be a little messy, but they are still trivially written in theory. For 5x5 or above, nothing can be done. NOTHING.
Do you see where this is going? Just wanting to solve a problem that is mathematically impossible to solve, where it has been proved to be the case. Well, the need or the desire is not sufficient. Sorry. This is not a question of MATLAB. It is a question of mathematical impossibility.
whos
Name Size Bytes Class Attributes A 6x6 8 sym A_evs 6x1 8 sym ans 6x1 8 sym cmdout 1x33 66 char gdsCacheDir 1x14 28 char gdsCacheFlag 1x1 8 double i 0x0 0 double managers 1x0 0 cell managersMap 0x1 8 containers.Map x 1x1 8 sym
It appears your matrix is A. That is a good name.
A
A =
And now you want to compute the eigenvalues of A, in an algebraic form, as a function of y AND of z. Sorry. Not gonna happen. EVER. Again, if you have a problem, all you can do is talk to either Abel of Ruffini. Maybe you can convince them that you really, very much need a solution, and that you need them to rewrite the proof. This may take a truly cosmic intervention though. Do you have the right connections for such a task? I certainly don't.
If you knew the values of both y and z. Well, then you can compute the numerical roots.
syms y z
vpa(eig(subs(A,[y,z],[3,17])))
ans =
But you need to see at that point, the result is just a numerical matrix. Now a rootfinder has no problem handling what is essentially a polynomial rootfinding problem.
In terms of your original matrix A though, there is NO numerical solution that can be applied. There is no magic. You really cannot do more than that. I'm sorry. Mathematics can be a cruel mistress at times.
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John D'Errico 2024 年 6 月 28 日

How much does it try to simplify the matrix? Not even at all hard, if it does anything at all. For example,...
syms x
A = sym(rand(5));
eig(sym(x*A))
ans =
So it is trivial to compute the eigenvalues of A (although computing them in an algebraic form is a mathematical impossibiity, as we should know.) Was it smart enough to factor out x, to know how the simple multiplicative factor x modifies the eigenvalues? Nope.
It is often the case that MATLAB almost looks like it is doing something, returning a rootof expression. For example:
syms a y
P = y^5 + a*y + 2;
ysol = solve(P == 0,y)
ysol =
It tells us that IF it could solve the problem, which it apparently does not fully understand that mathematically, it cannot, the result would be one of the 5 roots of that polynomial. It is a non-answer though.
Bastian Loß 2024 年 6 月 28 日

I have observed this behaviour as well, but this is essentially where my confusion comes from.
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.
Therefore, I thought Matlab would maybe use some "tricks" to transform my matrix beforehand.
Update: Or my Matrix maybe has some special properties that Matlab exploits to come up with an actual solution.

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