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A linear Trans. prob.

  1. Jul 23, 2012 #1
    Hello,
    I am a grad student preparing for a quals. I am using H. and Kunze book.
    the problem is:
    let V be a n-dim vector space over F. and let [itex]B[/itex]={[itex]a_1,a_2,..., a_n[/itex]} be an ordered bases for V.
    a- According to them 1, there is a unique Linear operator T on V such that

    [itex]Ta_i=a_{(i+1)}[/itex] , i=1,....,n. and [itex]Ta_n=0[/itex].
    what is the matrix [itex]A[/itex] of [itex]T[/itex] in the ordered bases B.
    b- prove that [itex]T^n=0, and\ \ \ T^{n-1}≠ 0[/itex].
    c- Let S be any linear operator on V such that [itex]S^n=0\ \ \ but\ \ \ S^{n-1}≠0[/itex]. Prove that there is an ordered bases [itex]B'[/itex] such that the matrix of [itex]S[/itex] in the bases [itex]B'[/itex] is the matrix [itex]A[/itex] of part (a).

    Solution Attempt.
    Obviously we have for (a)


    [itex]A= \begin{bmatrix}
    0 & 0&... & 0 &0\\
    1 & 0 & 0&....&0 \\
    0&1&0&.....&0\\
    .\\
    .\\
    0&0&....&1&0
    \end{bmatrix}[/itex]
    for (b) its obvious.
    my problem is with the last question. I tried to justfy it by two ways, the first one is to find an ivertible linear transformation [itex]U:V→V[/itex] such that [itex]S=UTU^{-1}[/itex], then we will be done and such a bases exists. The second way is that I am trying to show the follwing:
    there exist at least on vector in the bases [itex]B[/itex] such that [itex]S^ia_i≠0[/itex] for [itex]i=1,..,n-1.[/itex] and I am considering the set [itex]B'=[/itex]{[itex]a_i, Sa_i,S^2a_i,....,S^{n-1}a_i[/itex]}. Note if we proved [itex]B'[/itex] is a bases, then [itex]_{B'}=A[/itex].i.e. the matrix of [itex]S[/itex] relative to the bases [itex]B'[/itex] is A.
    Unfotunatly, I could not get to an end with both ways.
    Am I doing the right thing? Any suggestions?.
    Thank you in Advance.
     
    Last edited: Jul 23, 2012
  2. jcsd
  3. Jul 23, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hello Abuattallah! Welcome to PF! :smile:

    Hint: use the fact that Sn-1 ≠ 0 :wink:
     
  4. Jul 23, 2012 #3
    Thank you tiny-tim.
    This is my try:

    consider the base [itex]B'=[/itex]{[itex]a,Sa,....,S^{n-1}a[/itex]}, for some [itex]a[/itex] where [itex]S^ia≠0, \forall i≤n-1[/itex], now for the sake of simplicity lets call them as follow [itex]a=a_1, Sa=a_2,....,S^ia=a_{i+1},...,S^{n-1}=a_n[/itex],Consider the follwing

    [itex]c_1a_1+.......+c_na_n=0[/itex] ...... (1)
    Aplly [itex]S^{n-1}[/itex] to both sides of (1), we get
    [itex]c_1a_n=0→c_1=0[/itex],
    now apply [itex]S^{n-2}[/itex] to (1), we get
    [itex]c_2a_n=0→c_2=0[/itex], Continueing this way, we will get all [itex]c_i=0\ \ , \ \forall i=1,...,n[/itex], Hence the set [itex]B'[/itex] is linearly independent.
    Any comments on my method?

    Thanks in Advance.
     
  5. Jul 23, 2012 #4

    tiny-tim

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    Hi Abuattallah! :wink:

    Yes, that's right. :smile:

    Just a few "tweaks" needed …

    Start with "Since Sn-1 ≠ 0, there exists an a such that …"

    Then "So define the ordered set B' = {…}" (you can't call it a basis yet :wink:)

    And end "So B' is an independent ordered subset whose cardinality is the same as the dimension of V, and is therefore a basis for V"
     
  6. Jul 23, 2012 #5

    HallsofIvy

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    By the way, "bases" is the plural of "basis". We speak of a basis for a vector space.
     
  7. Jul 25, 2012 #6
    Thank you so much Tiny-Tim.
    and thanks for word correction, English is my second language ; ).
    You have all a wonderful day,
    Abuattallah
     
  8. Jul 25, 2012 #7

    HallsofIvy

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    Your English is far better than my (put just about any language here!).
     
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