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Math Help - Product of Linear Operators - A basic quesiton

  1. #1
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    Product of Linear Operators - A basic quesiton

    Let T_1, T_2,....T_n be linear operators on a vector space V.

    Let v\in V such that one of T_1(v),T_2(v),..,T_n(v) is 0.

    Is it ok to say T_1T_2....T_n(V)=0?
    Last edited by aman_cc; October 10th 2009 at 11:54 PM. Reason: latex tags error
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    Let T_1, T_2,....T_n be linear operators on a vector space V.

    Let v\in V such that one of T_1(v),T_2(v),..,T_n(v) is 0.

    Is it ok to say T_1T_2....T_n(V)=0?
    i think you meant T_1T_2....T_n(v)=0 not T_1T_2....T_n(V)=0? if so, the answer is no. for example let \{v_1,v_2 \} be a basis for V and define T_1(v_1)=0, \ T_1(v_2)=v_1, \ T_2(v_1)=v_2, \ T_2(v_2)=v_1.

    here's a much less trivial question: suppose that for every v \in V there exists 1 \leq i \leq n such that T_i(v)=0. would we necessarily have T_1T_2 \cdots T_n = 0?
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    Quote Originally Posted by NonCommAlg View Post
    i think you meant T_1T_2....T_n(v)=0 not T_1T_2....T_n(V)=0? if so, the answer is no. for example let \{v_1,v_2 \} be a basis for V and define T_1(v_1)=0, \ T_1(v_2)=v_1, \ T_2(v_1)=v_2, \ T_2(v_2)=v_1.

    here's a much less trivial question: suppose that for every v \in V there exists 1 \leq i \leq n such that T_i(v)=0. would we necessarily have T_1T_2 \cdots T_n = 0?
    Yes I meant T_1T_2....T_n(v)=0

    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 !
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    Yes I meant T_1T_2....T_n(v)=0

    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
    well, this is very different from your original question! suppose Tv=c_j v. then (T-c_kI)(v)=(c_j - c_k)v and so (T-c_1I) \cdots (T-c_kI)(v)=(c_j-c_k)(T-c_1I) \cdots (T-c_{k-1}I)(v) = \cdots

    i'm sure you can finish the proof now!
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    Quote Originally Posted by NonCommAlg View Post
    well, this is very different from your original question! suppose Tv=c_j v. then (T-c_kI)(v)=(c_j - c_k)v and so (T-c_1I) \cdots (T-c_kI)(v)=(c_j-c_k)(T-c_1I) \cdots (T-c_{k-1}I)(v) = \cdots

    i'm sure you can finish the proof now!
    Yes - I got it !! Thanks

    Now working on your non-trivial poser, let's see.
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    Quote Originally Posted by NonCommAlg View Post
    i think you meant T_1T_2....T_n(v)=0 not T_1T_2....T_n(V)=0? if so, the answer is no. for example let \{v_1,v_2 \} be a basis for V and define T_1(v_1)=0, \ T_1(v_2)=v_1, \ T_2(v_1)=v_2, \ T_2(v_2)=v_1.

    here's a much less trivial question: suppose that for every v \in V there exists 1 \leq i \leq n such that T_i(v)=0. would we necessarily have T_1T_2 \cdots T_n = 0?
    First - Thanks for your question. Here is my attempt (I hope I got it )
    Ans: Yes T_1T_2 \cdots T_n = 0 as I claim one of T_i= 0

    Re-phrasing your question.
    Let K_i be kernel of T_i
    You tell me  \bigcup K_i = V
    I claim this is possible only if one of K_i = V and hence corresponding T_i= 0

    If I show you that \forall i, if  K_i \neq V then  \bigcup K_i \neq V, I will be done. Correct?

    I will prove this by induction over i.  i=1 or i=2 are trivial I am omitting them.

    Let above be true for  i=n. I claim it is true for  i=n+1, for otherwise we have a situation where

     \bigcup K_i \neq V but  (\bigcup K_i)\bigcup K_{n+1} = V. Note also \forall i, K_i \neq V

    Under the above situation, I can find vectors
    a,b \in V such that a \in K_{n+1} and a \notin (\bigcup K_i) . Similarly b \notin K_{n+1} and b  \in (\bigcup K_i)

    Consider b, a+b, a+2n, ..., a+(n-1)b. No two of these can be in same K_i for i \in [1,n]. As other wise we will have a \in K_i, contrary to our claim.

    Therefor consider a+nb \in V. But, a+nb \notin  (\bigcup K_i)\bigcup K_{n+1}.

    A contradiction and hence we are done.

    @NonCommAlg: Please let me know if I am correct please.
    Thanks
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    Quote Originally Posted by aman_cc View Post

    Consider b, a+b, a+2n, ..., a+(n-1)b.
    considering the sentence after this i guess you meant: b, b+a, b+2a, \cdots , b+(n-1)a.



    a+nb \notin (\bigcup K_i)\bigcup K_{n+1}.
    why?
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  8. #8
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    Quote Originally Posted by NonCommAlg View Post
    considering the sentence after this i guess you meant: b, b+a, b+2a, \cdots , b+(n-1)a.




    why?

    Yes - I meant b, b+a, b+2a, \cdots , b+(n-1)a.. Sorry about that

    Any vector of the form b+ka can't belong to K_{n+1}. Because if it does => b+ka - (ka) = b \in K_{n+1}, which is a contradiction. Note:  a \in K_{n+1} =>  ka \in K_{n+1}

    So any vector of the form b+ka \in \bigcup K_i, i \in [1,n]

    Also b+k_1a, b+k_2a for two different k1,k2 cannot belong to same K_i, i \in [1,n]. Because if it does (b+k_1a) - (b+k_2a) = (k_1-k_2)a \in K_i. Which again is a contradiction as we have assumed a \notin \bigcup K_i, i \in [1,n]

    Now consider b+na. Sorry for typo here in the earlier post.

    By Pigeon-hole principle b+na can't belong to any of the K_i, i \in [1,n] .
    Also, b+na \notin K_{n+1}. as it is of the form b+ka

    Thus b+na \notin (\bigcup K_i)\bigcup K_{n+1}

    Thus (\bigcup K_i)\bigcup K_{n+1}\neq V
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  9. #9
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    Quote Originally Posted by aman_cc View Post
    Yes - I meant b, b+a, b+2a, \cdots , b+(n-1)a.. Sorry about that

    Any vector of the form b+ka can't belong to K_{n+1}. Because if it does => b+ka - (ka) = b \in K_{n+1}, which is a contradiction. Note:  a \in K_{n+1} =>  ka \in K_{n+1}

    So any vector of the form b+ka \in \bigcup K_i, i \in [1,n]

    Also b+k_1a, b+k_2a for two different k1,k2 cannot belong to same K_i, i \in [1,n]. Because if it does (b+k_1a) - (b+k_2a) = (k_1-k_2)a \in K_i. Which again is a contradiction as we have assumed a \notin \bigcup K_i, i \in [1,n]

    Now consider b+na. Sorry for typo here in the earlier post.

    By Pigeon-hole principle b+na can't belong to any of the K_i, i \in [1,n] .
    Also, b+na \notin K_{n+1}. as it is of the form b+ka

    Thus b+na \notin (\bigcup K_i)\bigcup K_{n+1}

    Thus (\bigcup K_i)\bigcup K_{n+1}\neq V
    it's correct now. good work! what you're saying is that for every 0 \leq i \leq n-1 there exists a "unique" 1 \leq j \leq n such that b + ia \in K_j. now b + na \notin K_{n+1} because a \in K_{n+1} and b \notin K_{n+1}.

    also b + na \notin K_j, for all 1 \leq j \leq n, because we already know that b + ia \in K_j for some 0 \leq i \leq n-1. so we'd have (n-i)a \in K_j and hence a \in K_j, 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!
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  10. #10
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    Quote Originally Posted by NonCommAlg View Post
    it's correct now. good work! what you're saying is that for every 0 \leq i \leq n-1 there exists a "unique" 1 \leq j \leq n such that b + ia \in K_j. now b + na \notin K_{n+1} because a \in K_{n+1} and b \notin K_{n+1}.

    also b + na \notin K_j, for all 1 \leq j \leq n, because we already know that b + ia \in K_j for some 0 \leq i \leq n-1. so we'd have (n-i)a \in K_j and hence a \in K_j, 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!
    Thanks very much. Really appreciate you being patient and helpful. I really enjoyed doing this question.

    Thanks again
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