# How to determine eigenvalues and eigenvectors?

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Md Ashikuzzaman on 13 Jul 2022
Edited: Issa on 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.

Bruno Luong on 13 Jul 2022
Edited: Bruno Luong on 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

### More Answers (2)

Issa on 13 Jul 2022
Edited: Issa on 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 on 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