Re: Are eigenvectors independent

In article <4orpboFg32u9U1@xxxxxxxxxxxxxx>,
=?ISO-8859-1?Q?Jos=E9_Carlos_Santos?= <jcsantos@xxxxxxxx> writes:
Luke Wu wrote:

I know about (and have seen proof of) the fact that eigenvectors
corresponding to distinct eigenvalues form a lin. independent seet.
(**)

But what about this type of situation:

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.

Derek Holt.

1 1

0 1

Then 1 is a root of the characteristic polynomial with multiplicity 2,
but you can't find two linearly independent eigenvectors with eigenvalue
1.

Is it true that these 2 eigenvectors along with the other 4 (each
related to a multiplicity 1 root) make an independent set?

*If* these two eigenvectors exist, then, yes, it is true, as Jannick
Asmus has already explained.

Best regards,

Jose Carlos Santos


.

Relevant Pages

  • Re: Checking names of eigenvalue "multiplicities"
    ... an eigenvalue e as the dimension of its space of generalised ... - perhaps algebraically closed in general, but Axler doesn't ... is usually called the "algebraic" multiplicity, ... the subspace spanned by the eigenvectors of given eigen- ...
    (sci.math)
  • Re: Checking names of eigenvalue "multiplicities"
    ... an eigenvalue e as the dimension of its space of generalised ... is usually called the "algebraic" multiplicity, ... the subspace spanned by the eigenvectors of given eigen- ... some power of T - lambda I, ...
    (sci.math)
  • Re: Are eigenvectors independent
    ... We are able to get 2 independent eigenvectors for this multiplicity 2 ... I think he is *assuming* the existence of two linearly independent ...
    (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 ... If is indep. ...
    (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)