AA =
How do I determine if a matrix is nilpotent using matlab?
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I tried using matrix manipulation to determine x which will determine whether A is nilpotent by using the following code:
A =[0 1 2 3 4 5 6 7;0 0 8 9 10 11 12 13;0 0 0 1 2 3 4 5; 0 0 0 0 6 7 8 9;0 0 0 0 0 1 2 3;0 0 0 0 0 0 4 5; 0 0 0 0 0 0 0 1;0 0 0 0 0 0 0 0];
B_A = zeros(8);
syms x
eqn = A.^x == B_A
solx = solve(eqn,x)
But I keep getting solx = Empty sym: 0-by-1
How do I get a solid solution for x? Or any other method to calculate the nilpotent will be helpful.
Réponse acceptée
Torsten
le 10 Avr 2017
1 vote
Your equation is wrong ; it must read
eqn = A^x==B_A
But you should not check for nilpotency this way.
In your case, it's obvious by inspection that A is nilpotent since it is upper triangular.
For the general case, I'd check whether A has only 0 as eigenvalue :
help eig
Or - provided n is small - just calculate A^n for an (nxn)-matrix A and see whether it's the null matrix.
Best wishes
Torsten.
11 commentaires
Amy Olivier
le 10 Avr 2017
Modifié(e) : Amy Olivier
le 10 Avr 2017
Torsten
le 10 Avr 2017
The first x for which A^x=0 cannot exceed n. So just test whether A^n = 0.
Best wishes
Torsten.
"For the general case, I'd check whether A has only 0 as eigenvalue"
Suppose we have the following matrix
z = sqrt(sym(2));
z1 = sqrt(sym(3));
AA = [-z, -z^2/z1; z1, z]
It is nilpotent and, as required, it has only zero as a repeated eigenvalue
AA*AA
eig(AA)
If we enter that matrix as a double
format short e
z = sqrt((2));
z1 = sqrt((3));
AAA = [-z, -z^2/z1; z1, z]
it is, perhaps surprisingly, nilpotent in double precision
AAA*AAA
any(ans(:))
But, its eigenvalues seem to be quite large. I mean, I'm not surprised the eigenvalues aren't exactly zero, but I was surprised they are on the order of 1e-8.
eig(AAA)
This example more inline with what I was expecting
z = 2;
z1 = 3;
AAA = [-z, -z^2/z1; z1, z]
AAA*AAA
eig(AAA)
In general, I'm curious about how large the eigenvalues can get, perhaps if the martrix itself is larger and/or if its index is higher. I'm also wondering if that first version of AAA has some property, or is member of a class of nilpotent matrices that share a common property, that will have a sensitive eig computation.
Any insights into any of this?
I don't think this is a question concerning nilpotency, but a more general question on how exact eigenvalues can be computed. Using a different algorithm gives a better result in this special case:
z = sqrt((2));
z1 = sqrt((3));
AAA = [-z, -z^2/z1; z1, z];
eig(AAA,eye(2),'qz')
z = 2;
z1 = 3;
AAA = [-z, -z^2/z1; z1, z];
eig(AAA,eye(2),'qz')
For sure, this is a question about computing eigenvalues.
Because AAA is not symmetric, the generalized eigenvalue solution will always use the qz algorithm, so it's really a question of what's the difference between eig(A) and eig(A,eye,alg). Is solving the former fundamentally different than the latter in some way? Is the latter always expected to be more accurate for matrices that have any repeated eigenvalues, or all repeated eigenvalues, or all repeated eigenvalues at 0?
z = sqrt((2));
z1 = sqrt((3));
AAA = [-z, -z^2/z1; z1, z];
e1 = eig(AAA)
e2 = eig(AAA,eye(2),'chol') % still uses qz
e3 = eig(AAA,eye(2),'qz')
isequal(e2,e3)
Is solving the former fundamentally different than the latter in some way? Is the latter always expected to be more accurate for matrices that have any repeated eigenvalues, or all repeated eigenvalues, or all repeated eigenvalues at 0?
Sorry, but I cannot answer this. Maybe Christine Tobler or Bruno Luong have experience with the two possible algorithms and their advantages/disadvantages for problems with repeated eigenvalues.
Another method to check 0 is eigen value of A is
A =[0 1 2 3 4 5 6 7;0 0 8 9 10 11 12 13;0 0 0 1 2 3 4 5; 0 0 0 0 6 7 8 9;0 0 0 0 0 1 2 3;0 0 0 0 0 0 4 5; 0 0 0 0 0 0 0 1;0 0 0 0 0 0 0 0];
isnilpotetent = ~isempty(null(A))
It is perhaps more "scalable" the EIG and friends
Torsten
le 14 Avr 2024
You had to check that 0 is the only eigenvalue A has ...
Bruno Luong
le 14 Avr 2024
In absolute my claim is not wrong I ener say "to check A is nilpotent". ;)
Here is another way to find the degree of Nilpotet matrix, ie smallest x such that A^x = 0
A =[0 1 2 3 4 5 6 7;
0 0 8 9 10 11 12 13;
0 0 0 1 2 3 4 5;
0 0 0 0 6 7 8 9;
0 0 0 0 0 1 2 3;
0 0 0 0 0 0 4 5;
0 0 0 0 0 0 0 1;
0 0 0 0 0 0 0 0];
J = jordan(A); % required symbolic toolbox
d0 = diag(J);
isnilpotent = all(d0 == 0);
if isnilpotent
d1 = diag(J,1); %
% length of the longest sequence of nonzero above the diagonal
% == jordan block size
j = diff([0; d1~=0; 0]); % positions where transition to/from 0 occur
l = diff(find(j)); % length of consecutive non-zeros
maxl = max(l);
x = sum([maxl 1]); % it return 1 if l/maxl is empty, add 1 otherwise
else
x = [];
end
x
Bruno Luong
le 15 Avr 2024
Modifié(e) : Bruno Luong
le 15 Avr 2024
It looks to me the EIG methods (both standarc and generalized) is difficult to be used for middle/large size nilpotetnt matrix. They seem to have difficulty to estimate 0 eigenvalues with huge errors.
Multiple order eigen values estimation is challenging to handle numerically.
n = 100; % 1000
A = diag(ones(1,n-1), 1);
[P,~]=qr(randn(size(A)));
B=P*A*P';
% B should be nimpotetnt, this is small
max(abs(B^n), [], "all")/norm(B)
% check eigen value of A
lambdaA = eig(A,eye(n),'qz','vector');
abs(lambdaA) % ok they are all 0Z
norm(lambdaA,'inf')
% Do the same for B
% check eigen value of B, orthogonal transform of A
lambdaB = eig(B,eye(n),'qz','vector');
sort(abs(lambdaB)) % However none of them is close to 0!!!
norm(lambdaB,'inf')
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