Effacer les filtres
Effacer les filtres

Calculating the eigenvalues of simple shapes

6 vues (au cours des 30 derniers jours)
John Bach
John Bach le 13 Mar 2023
Modifié(e) : John Bach le 25 Mar 2023
Hi there,
I've used the strel function to create a range of shapes and I would like to now calculate the eigenvalues of each shape although I am struggling to do this and would really appreciate any help regarding this.
Thank you in advance,
  1 commentaire
the cyclist
the cyclist le 13 Mar 2023
Modifié(e) : the cyclist le 13 Mar 2023
Do you have a reference for what the eigenvalue of a binary shape is? I did some googling of keywords, but didn't find something definite. (Maybe this is well known in image processing, but that is not my specialty.)
Are you stuck on the math of it, or the MATLAB coding? Have you written any code?

Connectez-vous pour commenter.

Réponse acceptée

Walter Roberson
Walter Roberson le 13 Mar 2023
M = double(strel('disk',5).Neighborhood)
M = 9×9
0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0
E = simplify(eig(sym(M)))
E = 
vpa(E)
ans = 
(the imaginary component is due to round-off error)
M2 = double(strel('octagon',12).Neighborhood)
M2 = 25×25
0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
E2 = simplify(eig(sym(M2)))
E2 = 
vpa(imag(E2))
ans = 
vpa(E2)
ans = 
  1 commentaire
Bjorn Gustavsson
Bjorn Gustavsson le 15 Mar 2023
The/One benefit of using svd instead of eig is that one get real singular values - which is not a guarantee with eig. Appart from that the soutions should be comparable/similar/identical.

Connectez-vous pour commenter.

Plus de réponses (1)

Bjorn Gustavsson
Bjorn Gustavsson le 13 Mar 2023
If you have a binary image then why not just run through the svd and see what you get:
I = zeros(256);
I(64:(64+128),64:(64+128)) = 1;
[U,S,V] = svd(I);
figure
subplot(1,2,1)
plot(diag(S))
subplot(1,2,2)
imagesc(U(:,1)*S(1,1)*V(:,1)')
% Or for a funnier example:
I = numgrid('B',258);
I = I(2:end-1,2:end-1);
[U,S,V] = svd(I);
subplot(2,2,1)
plot(diag(S))
subplot(2,2,2)
imagesc(U(:,1)*S(1,1)*V(:,1)')
subplot(2,2,2)
imagesc(U(:,1:4)*S(1:4,1:4)*V(:,1:4)')
subplot(2,2,2)
imagesc(U(:,1:16)*S(1:16,1:16)*V(:,1:16)')
You can also look at the individual eigen-images by something like:
imagesc(U(:,7)*V(:,7)')
HTH

Catégories

En savoir plus sur Linear Algebra dans Help Center et File Exchange

Produits

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by