Re: Are eigenvectors independent


mareg@xxxxxxxxxxxxxxxxxxxxxxxx wrote:
In article <4osd06Fg7ql7U2@xxxxxxxxxxxxxx>,
=?ISO-8859-1?Q?Jos=E9_Carlos_Santos?= <jcsantos@xxxxxxxx> writes:
mareg@xxxxxxxxxxxxxxxxxxxxxxxx wrote:

Say a matrix A has 5 eigenvalues (where one of them as multiplicity 2
as a root of characteristic equation of A).

We are able to get 2 independent eigenvectors for this multiplicity 2
eigenvalue.
Not necessarily. For instance, take A equal to

I think he is *assuming* the existence of two linearly independent
eigenvectors, rather than claiming that this is always true.

It is possible, but then why did he assume that one of the eigenvalues
has multiplicity 2 as a root of characteristic equation of A? There
would be no need to assume this, since it would follow from the
existence of 2 linearly independent eigenvectors

Well, no, from that it would only follow that the corresponding eigenvalue
had multiplicity at least 2.

My understanding of the question was: suppose, just to specific, that the
charcteristic equation is (x-1)^2 (x-2)(x-3)(x-4), and suppose in addition
that there are two linearly independent eigenvectors for the eigenvalue 1.
Can we conclude that there are five linearly independent eigenvectors,
and hence a basis of eigenvectors. The answer of course is yes.


Yeap. I am not the best at wording mathematical problems, but you're
spot on.
The "of course is yes" part was never touched on in my undergrad class.
I find it peculiar, since we spent a great deal of time discussing and
proving that eigenvectors corresponding to distinct/different
eigenvalues makes up a linearly independent set.

Since we didn't discuss the reasoning behind these "other cases" I
always thought it was just obvious becausee the distinct case can be
applied several times, but then I realized it couldn't.

.

Relevant Pages

  • Re: Are eigenvectors independent
    ... We are able to get 2 independent eigenvectors for this multiplicity 2 ... that there are two linearly independent eigenvectors for the eigenvalue 1. ...
    (sci.math)
  • Re: Are eigenvectors independent
    ... Say a matrix A has 5 eigenvalues (where one of them as multiplicity 2 ... We are able to get 2 independent eigenvectors for this multiplicity 2 ... I think he is *assuming* the existence of two linearly independent ... Then 1 is a root of the characteristic polynomial with multiplicity 2, ...
    (sci.math)
  • Re: Are eigenvectors independent
    ... Say a matrix A has 5 eigenvalues (where one of them as multiplicity 2 ... We are able to get 2 independent eigenvectors for this multiplicity 2 ... assuming is linearly independent, this means a=b=0. ... My intuition couldn't point me to this reasoning. ...
    (sci.math)
  • Re: Are eigenvectors independent
    ... Say a matrix A has 5 eigenvalues (where one of them as multiplicity 2 ... as a root of characteristic equation of A). ... We are able to get 2 independent eigenvectors for this multiplicity 2 ... assuming is linearly independent, this means a=b=0. ...
    (sci.math)
  • Re: eigenvector
    ... the only eigenvalues i can get are: -1 with a multiplicity of 2 and 0. ... with this in mind, i don't know how to proceed to get the eigenvectors. ... independent eigenvectors. ...
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