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Cyclic Decomposition Theorem

  1. Aug 18, 2011 #1
    I took an Intermediate Linear Algebra course all last year (two semesters worth) and we covered the CDT. My professor didn't teach it well, and I got my first B- in university because of it (didn't affect my GPA but still irritating).

    I didn't understand a lot of the canonical form stuff because I just straight up had a horrible prof who didn't allow us to grow a good conceptual idea of what the hell was happening.

    Then today i was reading the beginning of Artin's "Algebra" on groups, its it slapped me in the face and I was awestruck. He was talking about permutations, and the cyclic groups that may come out of them. My mind was blown as soon as I started thinking of the CDT.

    So, I'm just hoping for someone to explain it, and hopefully it matches my intuition.

    (I may be confusing CDT and RCF, i'm tired and can't think straight, thats why I'm waiting for someone first, unless they ask me for my explanation, then I will explain it tomorrow when I'm rejuvenated)
     
  2. jcsd
  3. Aug 19, 2011 #2
    Alright now that I have time to type. This won't be over the top rigourous, just trying to get my ideas onto 'paper'.

    Let [itex]V[/itex] be a vector space over a field [itex]\mathbb F[/itex] and let [itex]T[/itex] be a linear operator whose characteristic polynomial factors in [itex]k[/itex] irreducibles over [itex]\mathbb F[/itex]

    Basically, my intuition now tells me that we can find a basis so that the matrix representation of [itex]T[/itex] with respect to this new basis is 'prettier' and easier to work with. What I think the CDT tells us is that V can be broken down into [itex]k[/itex] [itex]T[/itex]-invariant subspaces that are disjoint. These subspaces are the null-spaces of each individual irreducible with [itex]T[/itex] substituted in. The basis we want is the union of the bases of these [itex]k[/itex] subspaces.

    I feel this, if this is correct (see how badly i was taught it hahahah), is analogous to the subsets a permutation creates. For example, let a permutation [itex]P:12345\to45132 [/itex]. Then we get the cyclic subsets [itex](143)(25)[/itex].

    Is this the proper intuition to have? Or am I way off the mark?
     
  4. Aug 19, 2011 #3

    micromass

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    No, I'm afraid this is not correct. Something that you describe is in fact the primary decomposition theorem (which is weaker than the cyclic decomposition theorem). For the primary decomposition theorem, we factor the minimal polynomial as

    [tex]p=p_1^{r_1}...p_n^{r_n}[/tex]

    where the [itex]p_i[/itex]are distinct. Then we indeed take the nullspace [itex]W_i[/itex] of [itex]p_i(T)^{r_i}[/itex]. Then we will indeed have [itex]V=W_1\oplus ... + W_n[/itex].

    However, the cyclic decomposition theorem is much harder and I don't think that such a trick will work there.
     
  5. Aug 19, 2011 #4
    Ooooooooh that makes sense.

    I don't currently have my Friedberg on me, so I wasn't able to look up the actual definitions.

    Well I'm definitely retaking that course because I really feel I missed out on a lot of good things. Ughzors.

    Thank you though!
     
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