eigenvalues and eigenvector manual calculation

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muhammad iqbal habibie
评论： Manoj Samal 2020-12-3
I have a matrix 2x2, for example A= [ 0.064911 3.276493; 3.276493 311.2073]. I would like to calculate the eigenvalues and eigenvectors. I have calculated the eigenvalues by manual and match it with matlab is match. the manual of eigenvalues :
eigenvalues were calculated by |A- λ * I|=0
so I received the eigenvalues (0.0304;311.2418). Now I am trying to calculated the eigenvectors that I found the way like this
B= eig(A) ; this is calculating the eigenvalues
(v,d)=eig(A)
I got v= (-0.9999 0.0105; 0.0105 0.9999) and d = ( 0.0304 0 ; 0 311.2418).
I would like to ask how to calculate manual of matrix v? Hope someone can help. Thank you.
  1 个评论
Manoj Samal
Manoj Samal 2020-12-3
By using (v,d)=eig(A) gives v= normalised eigen vector(not eigen vector) and d=eigen values
N11=1/sqrt(1^2+3^2+1^2)=1/sqrt11
N21=3/ sqrt(3^2+2^2+1^2)=1/sqrt14 and so on....

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采纳的回答

Torsten
Torsten 2018-11-19
编辑：Torsten 2018-11-19
By solving
(A-lambda1*I)*v1 = 0
and
(A-lambda2*I)*v2 = 0
You could use
v1 = null(A-lambda1*I)
and
v2 = null(A-lambda2*I)
to achieve this.
Best wishes
Torsten.
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muhammad iqbal habibie
muhammad iqbal habibie 2018-11-19
I try manual it is hard though
A- λ * I
so
[0.064911 3.276493; 3.276493 311.2073] - [0.0304 0; 0 311.2418]
= [0.034511 3.276493; 3.276493 -0.0345]
how must I suppose to be v = (-0.9999 0.0105; 0.0105 0.9999)
any suggestion for manual calculation?
Torsten
Torsten 2018-11-19
([0.064911 3.276493; 3.276493 311.2073] - [0.0304 0; 0 0.0304])*[v11; v21]=[0;0]
Solve for v1=[v11;v21] and normalize the vector to get the first column of v.
([0.064911 3.276493; 3.276493 311.2073] - [311.2418 0; 0 311.2418])*[v12;v22]=[0;0]
Solve for v2=[v12;v22] and normalize the vector to get the second column of v.

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更多回答（1 个）

Bruno Luong
Bruno Luong 2018-11-19
No cheating, this applies for 2x2 only
A= [ 0.064911 3.276493;
3.276493 311.2073];
lambda=eig(A); % you should do it by solving det(A-lambda I)=0
V = ones(2);
for k=1:2
B = A-lambda(k)*eye(size(A));
% select pivot column
[~,j] = max(sum(B.^2,1));
othercolumn = 3-j;
V(j,k) = -B(:,j)\B(:,othercolumn);
end
% Optional: Make eigenvectors l2 norm = 1
V = V ./ sqrt(sum(V.^2,1));
disp(V)
  2 个评论
muhammad iqbal habibie
muhammad iqbal habibie 2018-11-20
Is it possible from
A= [ 0.064911 3.276493;
3.276493 311.2073];
to v = (-0.9999 0.0105; 0.0105 0.9999) with eigenvalues (0.0304;311.2418) using excel?
Torsten
Torsten 2018-11-21
http://www.google.de/url?sa=t&rct=j&q=&esrc=s&source=web&cd=9&ved=2ahUKEwjF8uef_-TeAhUJ3qQKHZB3BSsQFjAIegQIBxAC&url=http%3A%2F%2Fwww-2.rotman.utoronto.ca%2F~hull%2Fsoftware%2FEigenvalue%26vector.xls&usg=AOvVaw2og1xfxU96ox_lDmx3k6GB

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