Let be linear operators on a vector space V.
Let such that one of is 0.
Is it ok to say ?
My book - says the following
If v is a eigenvector then one of
(T-c1I), (T-c2I),....,(T-ckI) sends v to 0.
where c1,c2,....ck are eigenvalues
I understand till this point.
Then it says, hence
[(T-c1I)(T-c2I)....(T-ckI)](v) = 0
This is where I got confused and asked the question above.
And let me think about your poser for a while please. Thanks again !
Ans: Yes as I claim one of
Re-phrasing your question.
Let be kernel of
You tell me
I claim this is possible only if one of and hence corresponding
If I show you that , if then , I will be done. Correct?
I will prove this by induction over . or are trivial I am omitting them.
Let above be true for . I claim it is true for , for otherwise we have a situation where
but . Note also
Under the above situation, I can find vectors
such that and . Similarly and
Consider . No two of these can be in same for . As other wise we will have , contrary to our claim.
Therefor consider . But, .
A contradiction and hence we are done.
@NonCommAlg: Please let me know if I am correct please.
Yes - I meant . Sorry about that
Any vector of the form can't belong to . Because if it does => , which is a contradiction. Note: =>
So any vector of the form
Also for two different cannot belong to same . Because if it does . Which again is a contradiction as we have assumed
Now consider . Sorry for typo here in the earlier post.
By Pigeon-hole principle can't belong to any of the .
Also, . as it is of the form
also for all because we already know that for some so we'd have and hence which is false.
Remark: as you might have noticed that your proof works only for vector spaces over fields with zero characteristic. this proof can be modified to work for all infinite fields. it is an interesting
question to see if there's any counter-example for vector spaces over "finite" fields!