Matching of eigenvalues of 2 matrices

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Omar Kamel
Omar Kamel 2019년 4월 29일
댓글: Omar Kamel 2019년 5월 3일
Hello Everybody,
Suppose I have 2 matrices with same size: . is a slight change of in the form of: . I do eigenvalue analysis on both matrices:
lambda_1 = eig(A_1);
lambda_2 = eig(A_2);
and want to do a comparison between eigenvalues of each matrix. How can I find the matching eigenvalue for the original , i.e. how can I identify each eigenmode from to its equivalent in ?

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Christine Tobler
Christine Tobler 2019년 4월 29일
With R2018a, there is a new function matchpairs which might be useful for this. Basically, it takes a matrix of similarities between two sets, and matches these sets up in pairs.
Here's an example:
>> X = randn(100);
>> d = eig(X);
>> d2 = eig(X + randn(100)*1e-2);
>> max(abs(d - d2))
ans =
17.9169
>> m = matchpairs(abs(d - d2.'), 1e-1); % The columns of m map the elements of d to those of d2
>> max(abs(d(m(:, 1)) - d2(m(:, 2))))
ans =
0.1569
The second input to matchpairs gives a cutoff: If two values have a difference larger than 1e-1, they would not be matched at all, and would show up as a new eigenvalue and a disappeared eigenvalues instead.
  댓글 수: 1
Omar Kamel
Omar Kamel 2019년 5월 3일
Thanks a lot Christine for your elaborate answer.

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추가 답변 (1개)

KSSV
KSSV 2019년 4월 29일
A1 = rand(2) ;
[v1,d1] = eig(A1) ;
Columns of v1 gives eigen vectors.........diagonal (diag(d1)) gives you eignvalues. To compare them use isequal. Or you may substract them and get the difference.
  댓글 수: 5
KSSV
KSSV 2019년 4월 29일
So sort them and keep them in order.....
Omar Kamel
Omar Kamel 2019년 4월 29일
It is not about sorting the list. It is about matching of the eigenvalues and eigenvectors to their equivalent in the other system, to know if any eigenmodes appeared or another eigenmodes disappeared.

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