I need to have eigenvalues of this matrix including symbols "m" and "n"

clear all
m=sym('m')
n=sym('n')
 A=sym([0 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m); 1/sqrt(3*n*m) 0 1/3 0 1/3 0 0 0 0; 1/sqrt(3*n*m) 1/3 0 1/3 0 0 0 0; 1/sqrt(3*n*m) 0 1/3 0 1/3 0 0 0 0; 1/sqrt(3*n*m) 1/3 0 1/3 0 0 0 0 0; 1/sqrt(3*n*m) 0 0 0 0 0 1/3 0 1/3; 1/sqrt(3*n*m) 0 0 0 0  1/3 0 1/3 0; 1/sqrt(3*n*m) 0 0 0 0 0 1/3 0 1/3; 1/sqrt(3*n*m) 0 0 0 0 1/3 0 1/3 0] )
eig(A)

[Walter Roberson]

 A=sym([0 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m); %length 9
        1/sqrt(3*n*m) 0 1/3 0 1/3 0 0 0 0;  %length 9
        1/sqrt(3*n*m) 1/3 0 1/3 0 0 0 0;    %length 8 !!
        1/sqrt(3*n*m) 0 1/3 0 1/3 0 0 0 0;  %length 9
        1/sqrt(3*n*m) 1/3 0 1/3 0 0 0 0 0;  %length 9
        1/sqrt(3*n*m) 0 0 0 0 0 1/3 0 1/3;  %length 9
        1/sqrt(3*n*m) 0 0 0 0  1/3 0 1/3 0; %length 9
        1/sqrt(3*n*m) 0 0 0 0 0 1/3 0 1/3;  %length 9
        1/sqrt(3*n*m) 0 0 0 0 1/3 0 1/3 0] )%length 9

[Walter Roberson]

You might have to wait quite a number of hours for the processing to finish, and you might need a lot of memory. Symbolic eigenvalues are typically quite expensive; it takes a fair while even just to do 5 x 5.

[neeraj kr]

I have faced the same problem that you talked about, just to solve a 4x4 matrix symbolic math is taking indefinite amount of time. Do not know what to do, can you suggest something to sort it out.

[Walter Roberson]

Sometimes what is most efficient is to create a second matrix that has a letter for each unique expression (but you can leave the 0's and 1's in place.) Take the eigenvalue of that, and then substitute for the values of the variables.

[Walter Roberson]

For example,

syms T q
A = [0 q q q q q q q q; %length 9
     q 0 T 0 T 0 0 0 0;  %length 9
     q T 0 T 0 0 0 0 0;    %length 8 !!
     q 0 T 0 T 0 0 0 0;  %length 9
     q T 0 T 0 0 0 0 0;  %length 9
     q 0 0 0 0 0 T 0 T;  %length 9
     q 0 0 0 0 T 0 T 0; %length 9
     q 0 0 0 0 0 T 0 T;  %length 9
     q 0 0 0 0 T 0 T 0]; %length 
 [Vs, Es] = eig(A);
syms m n
qs = 1./sqrt(3*m*n);
Ts = sym(1)/3;
V = subs(Vs, {q, T}, {qs, Ts});
E = subs(Es, {q, T}, {qs, Ts});

Answer 1

[Steven Lord]

For this particular case, it looks like the structure of the problem actually makes this not too bad to compute. I tried both with m and n as separate variables and with a substitution, z = m*n, since they never appear separately. Based on the pattern of A I assumed the missing element in the third row was a 0. You can see the pattern more easily once I replaced those longer expressions with just q and T (T for one third.)

syms z m n
q = 1./sqrt(3*z);
% q = 1./sqrt(3*m*n);
T = sym(1)/3;
A = [0 q q q q q q q q; %length 9
     q 0 T 0 T 0 0 0 0;  %length 9
     q T 0 T 0 0 0 0 0;    %length 8 !!
     q 0 T 0 T 0 0 0 0;  %length 9
     q T 0 T 0 0 0 0 0;  %length 9
     q 0 0 0 0 0 T 0 T;  %length 9
     q 0 0 0 0 T 0 T 0; %length 9
     q 0 0 0 0 0 T 0 T;  %length 9
     q 0 0 0 0 T 0 T 0]; %length 9
 [V, E] = eig(A);

But I agree with Walter. For a larger matrix A or for a more complicated pattern, the symbolic eigenvalue and eigenvector calculation may take a while and the expressions involved may become very, very long. For example, computing the eigenvalues and eigenvectors for a general 3-by-3 took about a tenth of a second and resulted in decently long expressions.

>> B = sym('B', [3 3]);
>> tic; [V, E] = eig(B); toc
Elapsed time is 0.095462 seconds.
>> strlength(char(V))
ans =
       30929

>> strlength(char(E))
ans =
        7301

Going to a 4-by-4 took a bit longer and resulted in much longer expressions.

>> B = sym('B', [4 4]);
>> tic; [V, E] = eig(B); toc
Elapsed time is 227.022771 seconds.
>> strlength(char(V))
ans =
     6586984

>> strlength(char(E))
ans =
      712104

It's dangerous to extrapolate based on only two data points, but if the 5-by-5 takes 2270 times as long as the 4-by-4 case, like the 4-by-4 case took 2270 times as long as the 3-by-3, it would take my machine about six days to generate that answer. Even if I just asked for E (not E and V) that would still run to about 7 million characters (again extrapolating based on the growth from the 3-by-3 to the 4-by-4 case.)

Consider substituting numbers for m and n (or z) and calling the numeric eig function instead. You can do this for many different numeric values of m*n in the time it would take you to solve the general problem symbolically and substitute.

Answer 2

[John D'Errico]

Actually, it is pretty easy, and the solution arrives almost immediately.

A=[0 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m) 1/sqrt(3*n*m);...
1/sqrt(3*n*m) 0 1/3 0 1/3 0 0 0 0;...
1/sqrt(3*n*m) 1/3 0 1/3 0 0 0 0 0;...
1/sqrt(3*n*m) 0 1/3 0 1/3 0 0 0 0;...
1/sqrt(3*n*m) 1/3 0 1/3 0 0 0 0 0;...
1/sqrt(3*n*m) 0 0 0 0 0 1/3 0 1/3;...
1/sqrt(3*n*m) 0 0 0 0 1/3 0 1/3 0;...
1/sqrt(3*n*m) 0 0 0 0 0 1/3 0 1/3;...
1/sqrt(3*n*m) 0 0 0 0 1/3 0 1/3 0];

First of all, this is not a problem where n and m enter independently. They always act together. We can arbitrarily substitute

u = 1/sqrt(3*n*m)

If a solution exists that we can find, then later on, substitute in the value for u. I'll even use v instead of 1/3 there, just to reduce my typing.

syms u
v = sym(1)/3;
A=[0 u u u u u u u u;...
u 0 v 0 v 0 0 0 0;...
u v 0 v 0 0 0 0 0;...
u 0 v 0 v 0 0 0 0;...
u v 0 v 0 0 0 0 0;...
u 0 0 0 0 0 v 0 v;...
u 0 0 0 0 v 0 v 0;...
u 0 0 0 0 0 v 0 v;...
u 0 0 0 0 v 0 v 0];

Now, the first question is, will we even be able to compute the eigenvalues of A analytically? That comes down to a rootfinding operation, on a degree 9 symbolic polynomial, with non-constant coefficients. I would immediately conjecture this will be difficult, since we know that is in general impossible for degree 5 polynomials or higher.

cp = charpoly(A)
cp =
[ 1, 0, - 8*u^2 - 8/9, -(16*u^2)/3, (32*u^2)/9 + 16/81, (64*u^2)/27, 0, 0, 0, 0]

Those are the coefficients, with the highest order term first. You can think of this as a polynomial in the unkown eigenvalue lambda.

The 4 zeros at the end mean we can factor out lambda^4. So there is an eigenvalue of multiplicity 4 at lambda==0. But that still leaves a degree 5 polynomial, so in general I still expect the problem to have no analytical solution.

syms lambda
reducedcp = sum(cp(1:6).*lambda.^(5:-1:0))
reducedcp =
lambda*((32*u^2)/9 + 16/81) - lambda^3*(8*u^2 + 8/9) + lambda^5 + (64*u^2)/27 - (16*lambda^2*u^2)/3

Can we solve it? Anything is possible, and sometimes we find an unexpectedly pleasant surprise.

EV = solve(reducedcp,lambda)
EV =
                                       -2/3
                                       -2/3
                                        2/3
 1/3 - (27*((32*u^2)/729 + 4/6561)^(1/2))/2
 (27*((32*u^2)/729 + 4/6561)^(1/2))/2 + 1/3

Add on the multiplicity 4 eigenvalue at lambda==0, and we have the eigenvalues.

EV = [EV;zeros(4,1)]
EV =
                                       -2/3
                                       -2/3
                                        2/3
 1/3 - (27*((32*u^2)/729 + 4/6561)^(1/2))/2
 (27*((32*u^2)/729 + 4/6561)^(1/2))/2 + 1/3
                                          0
                                          0
                                          0
                                          0

Now, you can even replace u with its value, in terms of n and m. As I said, the solution does not have m and n enter independently. They will always be together as a product.

Given the eigenvalues, now you can recover the associated eigenvectors. But you asked only for the eigenvalues.