Re: Are eigenvectors independent

On 08.10.2006 08:50, 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.
Is it true that these 2 eigenvectors along with the other 4 (each
related to a multiplicity 1 root) make an independent set? What is the
reasoning behind this?

Yes, it is true. Let l1, ..., l4 the pairwise different eigenvalues of
the 5x5 matrix A (l4 with multiplicity 2) and x1, ..., x5 the 5
eigenvectors where x4, x5 belong to eigenvalue l4.

Now test the vectors for linear independence: a1 x1 + ... + a5 x5 = 0
(**) with scalars ai. To show, e.g., a1 = 0, apply the product (A - l2
E)(A - l3 E)(A - l4 E) of matrices to (**) [E denotes the unit matrix].
Then proceed with a2, ...


By the way it generalizes to the result that the eigenspaces E_i to
different eigenvalues l_i of a matrix form a direct sum (+)_i E_i.

At first I thought it was obvious (by using theorem (**) 2 times), but
then I realized it wasn't.

Say we have 3 vectors a,b,c. If {a,b} is indep. and {b,c} is indep and
{a,c} is indep., how can we say that {a,b,c} is independent? (I
actually thought this was obviously true, but then realized it wasn't
true for many cases).

J.
.

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  • 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. ...
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