[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
> 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.
> 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);
Ok, I follow you up to this point.
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],
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 ->
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.
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