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Reduction of endomorphisms

  1. Dec 16, 2015 #1
    Hello, I am studying reduction of endomorphisms and I came across a theorem that I can't understand completely. It states that:

    Theorem: Let ##E## be a finite dimensional ##K## vector space, ##K## sub-field of ##\mathbb{C}##, and ##f## be an endomorphism of ##E##. Then, ##f## is diagonalizable if and only if there exists a polynomial ##P## of ##K[X]##, such that ##P(f) = 0##, and ##P## can be written as a product of polynomials of degree 1, and its roots have order of multiplicity 1.

    I understand the proof I have for the sufficient condition, but the proof for the necessary condition is hard to follow for me so I tried an alternate way. I would like you to tell me if this is correct please:

    ##\Leftarrow ## ) Assume that there exists distinct ##\lambda_i##'s for ##i = 1...p##, and a polynomial ##P## in the form ##P = a \prod_{i=1}^p (X - \lambda_i) ## such that ##P(f) = 0##.
    I want to show that ##E = \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}##, which is a necessary and sufficient condition of diagonalizability.
    We have that for any ##x\in E-\{0\}##, ## P(f)(x) = 0##. So there exists ##i \in \{ 1...p \}## such that ##f(x) = \lambda_i x##, and ##x## belongs to the eigenspace ##E_{f,\lambda_i}##. Therefore ##x\in \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}##, and ##E \subset \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}##. The other inclusion is trivial. So ##f## diagonalizable.
     
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  3. Dec 16, 2015 #2

    Samy_A

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    I may be missing the obvious, but I don't understand this step. How can it be that each element of E is an eigenvector?
     
  4. Dec 16, 2015 #3
    If ##P(f) = 0##, then ##P(f)(x) = 0 ## for all ##x\in E##.
    1. If ##x = 0## then ##x \in E \cap \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}##
    2. If ##x\neq 0##, then ## 0 = P(f)(x) = a\prod_{i = 1}^p (f(x) - \lambda_i x)##. So at least one term in the product is equal to 0. Therefore, there exists ##i## such that ##f(x) - \lambda_i x = 0##, which means that ##x## belongs to the eigenspace associated to ##f## and eigenvalue ##\lambda_i##.
     
  5. Dec 16, 2015 #4

    Samy_A

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    I'm very confused, and probably wasting your time, but let us take a trivial example.
    ##E={\mathbb C}²##, and ##f## the endomorphism represented (with the canonical basis) by the matrix ##\begin{pmatrix}1 & 0 \\
    0 & 2 \end{pmatrix}##.
    So ##f(a,b)=(a,2b)##, and ##P(X)=(X-1)(X-2)##.
    Take ##x=(1,1)##. ##f(x)=(1,2)##, so ##x## is not an eigenvector of f.
    But ##P(f(x))=f(f(x))-3f(x)+2x=f(1,2)-3(1,2)+2(1,1)=(1,4)-(3,6)+(2,2)=(0,0)##.
     
  6. Dec 16, 2015 #5
    Oooops, sorry it's me who's wasting your time, I realize I am completely confused with the notations. I need to re-do this. Sorry
     
  7. Dec 19, 2015 #6
    proofreading
     
    Last edited: Dec 19, 2015
  8. Dec 19, 2015 #7

    Samy_A

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    I'm somewhat troubled by this step, but maybe I'm wrong about the notation.
    If I understood it correctly, ##{Sp}(f)## is the set of all eigenvalues of ##f##.

    If that is indeed the case, ##n = \text{Card(Sp}(f))## is the number of eigenvalues of ##f##. You claim that ## \text{dim}(E) \le n ##. But that can't be true in general, because an endomorphism can have less distinct eigenvalues than the dimension of the vector space, and still be diagonalizable.

    I also don't immediately see why the linearly dependence of the family ## (x,f(x),f^2(x),...,f^n(x))## for all x implies that ## \text{dim}(E) \le n ##.
     
  9. Dec 19, 2015 #8
    Lol, this is exactly the reason why I deleted my post :-) I need to work more on this. Thank you for your help
     
  10. Dec 19, 2015 #9

    Samy_A

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    Ok, no problem. :)
     
  11. Dec 19, 2015 #10
    But I'm not giving up on this, I want to find the solution :-)
     
  12. Dec 19, 2015 #11
    Think I have it now.

    In a previous post, we said that ##P(f) = 0## implies that the eigenvalues of ##f## are among the zeros of ##P##.
    Then it had to be true that ## P(f) = 0 \iff Q(f) = 0 ##, where ##Q = \prod_{\lambda \in \text{Sp}(f)} (X - \lambda) ##.

    Now I want to show that ## E = \bigoplus_{\lambda\in\text{Sp}(f)} \text{Ker}(f-\lambda e) ##. The inclusion ##\supset## is trivial, and we now want to show ##\subset##. We need to prove that any ##x\in E## has a decomposition over the eigenspaces of ##f## in the form ##x = x_1 + ... + x_n ##, where ##x_i \in \text{Ker}(f-\lambda_i e) ##.

    That would be done if we could find ##n## polynomials ##Q_k## such that the endomorphism ##Q_k(f)## sends any ##x\in E## in the ##k##-th eigenspace and :
    ## 0 = Q(f) = e - \sum_{k=1}^n Q_k(f) ##.

    So the following constraints must be satisfied:
    1. ##Q_k(f) \in \text{Ker}(f-\lambda_k e) \iff f(Q_k(f)) = \lambda_k Q_k(f) \iff X(f) \circ Q_k(f) = (\lambda_k e)(f)\circ Q_k(f) \iff ((X-\lambda_k ) Q_k)(f) =0 ##
    2. Existence of a non-zero constant ##\beta## such that ## Q = \beta \ (1 - \sum_{k=1}^n Q_k)##, and the ##Q_k##'s have the same degree as ##Q##.
    So if ##Q_k## has the form ##Q_k = \alpha (X-\mu) \prod_{i\neq k} (X-\lambda_i)##, where ##\alpha,\mu## are constants to be determined, then the first constraint is satisfied, and ##Q_k## has the same degree as ##Q##. For the second constraint, we must satisfy that ## 1 - \sum_{k=1}^n Q_k ## has the same roots as ##Q##. So it seems logic to determine ##\alpha,\mu## such that ##Q_k(\lambda_i) = \delta_{ik} \iff \alpha = \frac{1}{(\lambda_k-\mu) \prod_{i\neq k} (\lambda_k - \lambda_i)}## and ##\mu## is not an eigenvalue of ##f##.

    Finally, there exists a constant ##\beta \neq 0## such that ## Q = \beta (1 - \sum_{k=1}^n \frac{X-\mu}{\lambda_k -\mu} \prod_{i\neq k} \frac{ (X-\lambda_i)}{ (\lambda_k - \lambda_i)})##, and for any ##x\in E##, ##P(f)(x) = 0 \iff Q(f)(x) = 0 \Rightarrow x \in \bigoplus_{\lambda\in \text{Sp}(f)} \text{Ker}(f-\lambda e) ##, which proves that ##f## is diagonolizable.

    Is it correct now ?
     
  13. Dec 20, 2015 #12

    Samy_A

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    Looks correct to me. I'm a little lost in your point 1) concerning the polynomials ##Q_k##, but the formula for these polynomials fits the bill.
     
    Last edited: Dec 20, 2015
  14. Dec 20, 2015 #13
    Finally! That made me sweat ;-) Thank you very much Samy for your patience
     
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