Eigen Values and Eigen Vectors Matlab

问题:

I have a matrix A

A = [ 124.6,95.3,42.7 ; 95.3,55.33,2.74 ; 42.7,2.74,33.33 ]

The eigenvalues and vectors:

[V,D] = eig(A) 

How do I show the eigenvalues are mutually perpendicular? I've tried that if the dot product of the eigenvalues are zero, this demonstrates they are mutually perpendicular, but how would you compute this in MATLAB? I tried the following code

transpose(diag(D)) * diag(D)  %gives 4.1523e+04

Also, how can I verify the definition of eigenvalues and vector holds:

A e_i - L_i e_i = 0 

The above equation: for i equal 1 to 3. For a real, symmetric matrix all eigenvales are positive numbers and eigenvectors for a basis for the matrix

I tried the following code, but it doesn't seem to give me 0. Any ideas?

A*V(1) - D(1)*V(1)

回答1:

To show orthogonality

>> V'*V-eye(size(V))

ans =

   1.0e-15 *

    0.2220    0.1110    0.2498
    0.1110   -0.4441    0.1388
    0.2498    0.1388    0.4441

To show that the definition of the eigendecomposition is satisfied,

>> A*V - V*D

ans =

   1.0e-13 *

    0.4086    0.0400    0.8527
    0.3908    0.0355    0.5684
    0.1954    0.0355         0

The results won't be exactly zero, because digital computers don't do exact math, but you can see that they're pretty close.

回答2:

To compute all dot products between columns of V:

M = squeeze(sum(bsxfun(@times, conj(V), permute(V, [1 3 2]))));

The columns of V (eigenvectors) will be orthogonal if the above M is diagonal.