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Generalized eigenvectors

  1. Nov 16, 2011 #1

    Geofleur

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    Please note: Below, I keep trying to put [ capital B ] but it gets turned into !!

    In Dennery and Krzywicki, they give an example of how to put a matrix in Jordan canonical form (pp. 167-170). They start with a 4x4 matrix [A] that looks kind of messy and transform it to a quasi-diagonal form [A]' = [B^(-1)] [A] that is less full (I'm distinguishing between operators A and their matrix representations [A]). So far so good.

    They go on to say that vectors in the null space of (A-2E)^3 are also in the null space of ([A]' - 2[E])^2. This, they say, shows that A has no generalized eigenvectors of rank 3. They seem to be assuming that if ([A]'-2[E])^2 [v]' = 0 then (A-2E)^2 |v> = 0 (|v> is a vector without reference to a particular basis, and [v]' = [B^(-1)] [v]). Then they would have

    (A-2E)^3 |v> = 0 implies that (A-2E)^2 |v> = 0

    and this statement does indeed seem incompatible with the existence of a generalized eigenvector of rank 3 for A.

    Now for my big confusion:

    If ([A]'-2[E])^2 [v]' = 0 is just a particular representation of (A-2E)^2 |v> = 0, then it would seem that so is ([A]-2[E])^2 [v] = 0. But

    ([A]'-2[E])^2 [v]' = ([B^(-1)] [A] - 2[E])^2 [B^(-1)] [v] =

    [B^(-1)]^2 ([A] - 2[E])^2 ^2 [B^(-1)] [v] =

    [B^(-1)]^2 ([A] - 2[E])^2 [v] = 0.

    The operators ([A]'-2[E])^2 and ([A] - 2[E])^2 do not seem to have to same vectors in their nullspaces. There is a factor of that would have to be the identity matrix there in the last equation. So what, then, does it mean to talk about *the* nullspace of (A-2E)^2? What would particular representations of the operator look like?
     
  2. jcsd
  3. Nov 16, 2011 #2

    lurflurf

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    That should be
    [ A ]' = [ B-1 ] [ A ] [ B ]

    [B] comes out looking like due to forum tags

    it is obvious that ([ A ]'-2[ E ])2 and ([ A ] - 2[ E ])2 do have to same vectors in their nullspaces as

    [ B ]([ A ]'-2[ E ])2=([ A ] - 2[ E ])2[ B ]
     
    Last edited: Nov 16, 2011
  4. Nov 16, 2011 #3

    Geofleur

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    I don't see how that last statement is true. If I replace [A]' with the similarity transformation and plug it into the equation on the left hand side, I don't see how the right hand side results. There would be a factor of the inverse of B squared, for instance, no?

    Also, I don't see how equality on multiplying by the inverse of B implies equality of the nullspace vectors.
     
  5. Nov 16, 2011 #4

    Geofleur

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    Wait. Actually I do see it now. If you write the squared factors out explicitly and replace [A]' with the similarity transformation in each factor, the two operators I mentioned can be seen to have the same nullspace vectors. Thanks!

    I should mention that before, I was pushing operators through several terms without realizing it. Writing out the factors explicitly kept me from doing that.
     
  6. Nov 16, 2011 #5

    lurflurf

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    Oops your right that should have been

    [ B ]([ A ]'-2[ E ])2=([ A ] - 2[ E ])2[ B ]
     
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