A serious problem when calculating eigenvectors of a matrix with the function eig

Hi all,

I calculated the eigenvalues of A=[1 2;3 4] as [vec,val]=eig(A) and I found:vec=[-0.8246 -0.4160;0.5658 -0.9094] and val=[-0.3723 0;0 5.3723].

With Mathematica and Maple, the eigenvalues of A are the same as Matlab, but vec=[0.45743 1.0000;-1.4574 1.0000] (the same in maple and mathematica) is totaly different than Matlab.

It is a serious problem.

Thank you

[Boughrara kamel]

It is a serious problem. Is there another way to calculate eigenvectors with Matlab or I miss some things?

Answer 1

[Sebastian Castro]

The results are actually the same, but (not knowing anything about Maple and Mathematica) the formatting is different. It looks like

• The MATLAB vectors are normalized while the Maple/Mathematica ones aren't -- instead, they have one element with a value of 1.0 and the other one is relative to that.
• The MATLAB vectors are expressed in columns while the Maple/Mathematica are in rows.

To test this out, I did the following:

>> v1_maple = [0.45743 1];
>> v1_maple/norm(v1_maple)

ans =

      0.4160    0.9094

>> v2_maple = [-1.4575 1];
>> v2_maple/norm(v2_maple)

ans =

     -0.8246    0.5657

Compare those results with the columns of the MATLAB results and there you go! Well, one says 0.5657 and the other says 0.5658, but that might be a display/rounding issue.

- Sebastian

[Boughrara kamel]

Thank you for the answer, but how can I get with matlab the eigenvalues and vectors without the normalisation ?

[Sebastian Castro]

If you really want them to not be normalized, then just divide them by the element you'd like to set to 1. For example,

>> V = [-0.8246; 0.5658];
>> V/V(2)

ans =

     -1.4574
      1.0000

- Sebastian

[Greg Obi]

hey can you expound more on how you can get the results to be unormalized. I'm having the same issue and I'm getting closer to understanding it.

[Steven Lord]

By the definition, if V is an eigenvector of a matrix A with eigenvalue d, then A*V = V*d.

Now let's consider a new vector W = 2*V. Is W an eigenvector of A, and if so what's its eigenvalue?

A*W = A*2*V   % Since W = 2*V
A*2*V = 2*A*V % Since scalars commute with matrices
2*A*V = 2*V*d % Since V is an eigenvector of A with eigenvalue d
2*V*d = W*d   % Since 2*V = W

Therefore A*W = W*d. That means W is also an eigenvector of A and it has the same eigenvalue d as V does.

You can generalize this argument to any non-zero multiple of the eigenvector.

A = rand(5);
[V, D] = eig(A);
v1 = V(:, 1);
d1 = D(1, 1);

% v1 is an eigenvector of A with eigenvalue d1
shouldBeSmall1 = A*v1-v1*d1

% So is pi*v1
shouldBeSmall2 = A*(pi*v1)-(pi*v1)*d1

% We don't need to restrict ourselves to real number either
% (3+4i)*v1 is an eigevector as well
c1 = (3+4i)*v1;
shouldBeSmall3 = A*c1-c1*d1

In Sebastian's example, the chosen non-zero multiplier was (1/V(2)). You can choose a different multiplier if it's appropriate for the physical application the matrices whose eigenvalues you're computing are modeling.