Re: Are eigenvectors independent

In article <1160290201.047555.48670@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Luke Wu <LookSkywalker@xxxxxxxxx> 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?

Say v1 and v2 are eigenvectors corresponding to 2, and w1,w2,w3,w4 are
the others. Then for any scalars a and b, (a*v1 + b*v2) is either 0 or an
eigenvector corresponding to 2. If

a*v1 + b*v2+ c1*w1 + ... + c4*w4 = 0

then


a*v1 + b*v2 = -(c1*w1 + ... + c4*w4).

But the only way you can have an element of the eigenspace E_2 be a
linear combination of eigenvectors corresponding to other eigenvalues
is if the element is 0; therefore a*v1+b*v2 = 0, and since we are
assuming {v1, v2} is linearly independent, this means a=b=0. Now we
just have

c1*w1 + ... + c4*w4 = 0

and we deduce c1=c2=c3=c4=0, so they are linearly independent.


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?

No.

a = (1,0)
b = (0,1)
c = (1,1)

in R^2.

But you don't need this for the argument above.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

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