# [Maxima] eigenvalue, eigenvector

Raymond Toy (RT/EUS) raymond.toy at ericsson.com
Thu May 29 17:02:02 CDT 2008

```ahmet alper parker wrote:
>
>
> On Wed, May 28, 2008 at 9:24 PM, ahmet alper parker <aaparker at gmail.com
> <mailto:aaparker at gmail.com>> wrote:
>
>     I am trying to do a modal analysis with K the stiffness matrix and M
>     the mass matrix and trying to find K*lamda=w^2*M*lamda. Looking at
>     the matlabs help files, I discovered that I can use (M^-1.K) to
>     calculate the eigenvalues and eigenvectors with the standard
>     functions. But when I tried to find (according to my knowledge) the
>     eigenvectors, The only method I know did not worked for the
>     eigenvectors.
>     For example
>     writing w^2 in the equation and solving for lamda, the lamda is a
>     vector and as far as I know I have to enter 1 to at least one of
>     them, since they are not independent. Then I solve the others from
>     the matrix equation.
>
>
>     K:matrix([(12*E*I)/L^3,0,0],[0,(3.555555555555555*E*I)/L^3,-(3.555555555555555*E*I)/L^3],[0,-(3.555555555555555*E*I)/L^3,(7.111111111111111*E*I)/L^3]);
>     M:matrix([(m*L)/2,0,0],[0,(5*m*L)/4,0],[0,0,(3*m*L)/2]);
>     J:K-w^2*M;
>     solve([determinant(J)=0;], [w]);
>
>     I got eigenvalues correctly.
>     subst((8*sqrt(sqrt(34)+8)*sqrt((E*I)/m))/(3*sqrt(15)*L^2), w, J);
>     t:%;

Ok, I follow you up to this point.

>     [lamda1=0,lamda2=(48*%r1^2-2046*%r1)/(325*%r1+410),lamda3=%r1];

I assume this is what you get from solving the equation J with the value
of w given above.

It seems this might be a bug in linsolve.  (Well, I used linsolve to get
the same expression as you did above.)

However, note that K can be simplified to be a be

K1 : matrix([12*c,0,0],[0,3.555555555555555*c,-3.555555555555555*c],
[0,-3.555555555555555*c,7.111111111111111*c]);

where c = E*I/L^3 and M can be simplified to be

M1 : matrix([ml,0,0],[0,5*ml/2,0],[0,0,3*ml]);

where ml = m*L.  Then

J1 : K1-w^2*M1 ->

matrix([12*c-ml*w^2,0,0],
[0,3.555555555555555*c-5*ml*w^2/2,-3.555555555555555*c],
[0,-3.555555555555555*c,7.111111111111111*c-3*ml*w^2]);

If we then solve this for w, and substitute a value back into J1, and
ask linsolve for the solution, we get:

[f1 = 0,f2 = -(sqrt(34)*%r4-2*%r4)/5,f3 = %r4]

so we get nice linear relationship between the eigenvector [f1,f2,f3]
and the arbitrary parameter %r4.

Ray
```