Let be linear operators on a vector space V.

Let such that one of is 0.

Is it ok to say ?

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- Oct 11th 2009, 12:53 AMaman_ccProduct of Linear Operators - A basic quesiton
Let be linear operators on a vector space V.

Let such that one of is 0.

Is it ok to say ? - Oct 11th 2009, 01:34 AMNonCommAlg
- Oct 11th 2009, 01:44 AMaman_cc
Yes I meant

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 ! - Oct 11th 2009, 02:06 AMNonCommAlg
- Oct 11th 2009, 09:03 PMaman_cc
- Oct 12th 2009, 01:40 AMaman_cc
First - Thanks for your question. Here is my attempt (I hope I got it :))

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.

Thanks - Oct 12th 2009, 04:25 AMNonCommAlg
- Oct 12th 2009, 05:17 AMaman_cc

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

Thus

Thus - Oct 12th 2009, 05:52 AMNonCommAlg
it's correct now. good work! what you're saying is that for every there exists a "unique" such that now because and

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! (Evilgrin) - Oct 12th 2009, 07:43 AMaman_cc