eigenvalues and eigenvector manual calculation

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.

[Manoj Samal]

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....

Answer 1

[Torsten]

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.

[muhammad iqbal habibie]

lambda1 means eigenvalues1?

[Torsten]

Yes, "lambda" is the λ in

|A- λ * I|=0

[Bruno Luong]

This is still cheating: NULL calls SVD so there is a eigen-value calculation under the hood.

[muhammad iqbal habibie]

Oke I try using this v1 = null(A-lambda1*I) . only appear v1= 2×0 empty double matrix not the results as v.

[Torsten]

null(A) determines the nullspace of the matrix A.

[muhammad iqbal habibie]

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]

([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.

Answer 2

[Bruno Luong]

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)

[muhammad iqbal habibie]

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]

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