Porque el comando Autovalores no funciona con la Matriz {{0,1,0},{0,0,1},{2,-5,4}}?

Julio Morales shared this question 1 month ago
Answered

Comments (13)

photo
1

Hi, ¿que es el comando Autovalores ?

photo
1

Me imagino que se refiere al comando Eigenvalues().

Este comando es de la vista CAS. Se tiene que usar ahí. De otra manera no funciona.

Saludos.

photo
1

Perdón. Quise decir que el comando Eigenvectors (Autovectores) no funciona con dicha matriz

photo
1

parece un bug

estas instrucciones son muy recientes y provienen de conseguir la matriz diagonal. supongo que al no ser diagonalizable porque solo hay dos autovectores (uno por autovalor y uno de ellos es doble) se produce una salida de indefinido en lugar de decir cuales son los autovectores.


supongo que lo remediarán o que pondrán en la ayuda que solo da los autovectores en caso de existir la matriz diagonal. probaré la hipotesis anterior buscando una matriz no diagonalizable que responda bien al comando. supongo que no la encontraré

PD: efectivamente no me da los autovectores si la matriz no es diagonalizable.

mientras se podría dar un rodeo con la instrucción JordanDiagonalisation()

photo
1

Gracias

photo
photo
1

Bernard says:

1 is eigenvalue of multiplicity 2, but the eigenspace is of dimension 1, hence the matrix is not diagonalizable.

Try

JordanDiagonalisation( {{0,1,0},{0,0,1},{2,-5,4}})

photo
1

Hi Michael!


Yes, a matrix can be diagonalized iff the geometric multiplicity=algebraic multiplicity for each eigenvalue.

BUT:


eigenvectors exists also for not diagonalizable matrices like the given example!


The command "eigenvalue" delivers the eigenvalues with multiplicity and I would expect, that the command "eigenvector" also indicates the eigenvector(s) for each eigenvalue.

This command should also work for not diagonalizable matrices, because these matrices could also have eigenvectors.

The "concept eigenvector" is a more general concept than "diagonalizable".

May be that can be changed.


Greetings

mire2

photo
1

maxima says


/GoFDWhAAAAABJRU5ErkJggg==

I'd like to know the multiplicity of eigenvalues also

photo
1

Salut mathmagic!


Hmm, I'm not sure if I understand your wish correct, but look at Juans posted picture.

4c324c6b87d06fe62d0ffd68f62f6b04Like in maxima you can see, that the eigenvalue 2 has multiplicity 1 and the eigenvalue 1 has multiplicity 2.

That's the reason why the eigenvector-command in GG not works for not diagonalizable matrices, because the algebraic multiplicity of the eigenvalue 1 is 2, but the geometric multiplicity is 1 (the eigenspace of the eigenvalue 1 is span{(1,1,1)^t} and for the eigenvalue 2 it"s span{(1,2,4)^t})


And I think, that the eigenvector-command in GG should output these eigenvectors (1,1,1)^t and (1,2,4)^t as you can see in maxima.


Greetings

mire2

photo
1

excuse me, sorry,I should have seen it in old applet I did in past

photo
1

/iF9BjKWbviQAAAABJRU5ErkJggg==

fallaría si un autovalor es 0. en ese caso habría que seguir otro camino que evite la división, pero no sería dificil

photo
1

Hi mathmagic!


Sorry, but I don't understand your screenshot and the Google-Translation into German is funny, rather unintelligible.

There is no eigenvalue 0 and I don't understand the calculations from line 3ff and what they should show.

Maybe I'm to stupid or Google-Translation filtered the main informations. oO


c0807e8470d08b3c1e946cb7b8091bf7

Kind regards

mire2

photo
1

6 shows the eingenvectors and 7 shows the vector of Ker((M-λ I)²) for jordan matrix


the latin langs have time for verbs that are very hard for translating

if there is an eigenvalue=0 the method fails


3 is P*D. being M=P*D*P⁻¹

© 2018 International GeoGebra Institute