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How to determine eigenvalues and eigenvectors?

11 vues (au cours des 30 derniers jours)
Md Ashikuzzaman
Md Ashikuzzaman le 13 Juil 2022
Modifié(e) : Abderrahim. B le 13 Juil 2022
I have two matrices for example A and B. A=[3,9;3,5] and B=[2,0;0,8].
They are part of an eigenvalue problem of the form: (A-(lambda)B)x=0.
How do I find the eigenvalues and vectors using matlab? Please solve this problem using values and sharee the code from your monitor if possible.

Réponse acceptée

Bruno Luong
Bruno Luong le 13 Juil 2022
Modifié(e) : Bruno Luong le 13 Juil 2022
A=[3,9;3,5]
A = 2×2
3 9 3 5
B=[2,0;0,8]
B = 2×2
2 0 0 8
[V lambda] = eig(A,B,'vector')
V = 2×2
1.0000 -1.0000 0.2074 0.4018
lambda = 2×1
2.4332 -0.3082
% here is the first eigen vector with lambda(1) the corresponfing eigen
% value
x1 = V(:,1)
x1 = 2×1
1.0000 0.2074
(A - lambda(1)*B)*x1 % small but not 0 due to finite precision floating point
ans = 2×1
1.0e-15 * 0.2220 -0.4441
% second eigen vector and second eigen value lambda(2)
x2 = V(:,2)
x2 = 2×1
-1.0000 0.4018
(A - lambda(2)*B)*x2 % small but not 0 due to finite precision floating point
ans = 2×1
1.0e-15 * 0.8882 0.4441

Plus de réponses (2)

Abderrahim. B
Abderrahim. B le 13 Juil 2022
Modifié(e) : Abderrahim. B le 13 Juil 2022
Use eig
A = [3,9;3,5];
[eVecs, eVals] = eig(A)
eVecs = 2×2
-0.9026 -0.8196 0.4304 -0.5729
eVals = 2×2
-1.2915 0 0 9.2915
Eigenvalues are the diagonal elements of eVals. To get them use diag
eValues = diag(eVals)
eValues = 2×1
-1.2915 9.2915

Chunru
Chunru le 13 Juil 2022
% doc eig for more details
A=[3,9;3,5]
A = 2×2
3 9 3 5
B=[2,0;0,8]
B = 2×2
2 0 0 8
[vA, dA] = eig(A)
vA = 2×2
-0.9026 -0.8196 0.4304 -0.5729
dA = 2×2
-1.2915 0 0 9.2915
[vB, dB] = eig(B)
vB = 2×2
1 0 0 1
dB = 2×2
2 0 0 8

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