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How to determine eigenvalues and eigenvectors?

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Md Ashikuzzaman
Md Ashikuzzaman am 13 Jul. 2022
Bearbeitet: Abderrahim. B am 13 Jul. 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.

Akzeptierte Antwort

Bruno Luong
Bruno Luong am 13 Jul. 2022
Bearbeitet: Bruno Luong am 13 Jul. 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

Weitere Antworten (2)

Abderrahim. B
Abderrahim. B am 13 Jul. 2022
Bearbeitet: Abderrahim. B am 13 Jul. 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 am 13 Jul. 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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