Re: Are eigenvectors independent

In article <1160342412.093510.195240@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Luke Wu <LookSkywalker@xxxxxxxxx> wrote:

Nice. My intuition (or lack of it) couldn't point me to this reasoning.
Thanks.

No problem.

Perhaps the best way to see it in full generality is to take a slight
generalization of the statement "eigenvectors corresponding to
distinct eigenvalues are linearly independent":

THEOREM. Let c1,...,ck be pairwise distinct scalars, V a vector space,
and T a linear operator on V. Let v1,...,vk be vectors such that T(vi)
= ci*vi for i=1,2,...,k. If

(*) v1 + ... + vk = 0

then v1=v2=...=vk=0.

Proof. We proceed by induction on k. If k=1 then the result is
trivial. Assume then that the result holds for k, and let us consider
the situation with k+1 vectors. Notice it is the same proof as the one
for the statement I quoted above. Multiply (*) by c1. We get

c1*v1 + c1*v2 + ... + c1*v(k+1) = 0.

Now apply T to (*) to get

c1*v1 + c2*v2 + ... + c(k+1)*v(k+1) = 0.

Subtracting this from the one above we get

(c1-c2)*v2 + ... + (c1-c(k+1))*v(k+1)0

Now, notice that if r is any scalar, then T(r*vi) = rT(vi)=r(ci*vi) =
ci*(r*vi). Thus, applying the theorem for k vectors, using (c1-c2)*v2,
...., (c1-c(k+1))*v(k+1), we conclude that

(c1-ci)*vi = 0 for i=2,...,(k+1).

Now, c1-ci is nonzero since by assumption all the ci are pairwise
distinct. Therefore, the only way we can have (c1-ci)*vi = 0 is if
vi=0 for i=2,...,(k+1). Going now back to (*), we have

0 = v1 + v2 + ... +v(k+1) = v1,

so v1=v2=....=v(k+1)=0, as claimed. QED

Now we can get what you want:

THEOREM. Let T be a linear operator on V, let c1,...,ck be pairwise
distinct eigenvectors. If v_{i1},...,v_{i,n(i)} are linearly
independent eigenvectors corresponding to ci, i=1,...k, then
{v_{rs}}, 1<= r <= k, 1<= s <= n(r), is linearly independent.

Proof. Consider a linear combination of the v_{rs},

Sum a_{rs}v_{rs} = 0.

Let w_i = Sum a_{is}*v_{is} 1<= s <= n(i).

Then each of w_1,...,w_k satisfies T(w_i)=ci*w_i, and we have
w_1+...+w_k=0. By the theorem above, w_i = 0 for i=1,...,k.

Now we have

0 = w_i = a_{i1}v_{i1} + ... + a_{i,n(i)}v_{i,n(i)}.

Since by assumption v_{i1},...,v_{i,n(i)} are linearly independent,
this means a_{i1}=...=a_{i,n(i)}=0. This holds for i=1,...,k, proving
that {v_{rs}} is linearly independent. QED

So you can find bases for the distinct eigenvalues, put them together,
and be sure you have a linearly independent set.

--
======================================================================
"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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