Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

About the invariance of similar linear operators and their minimal polynomial

  1. Jun 20, 2009 #1
    About the invariance of similar linear operators and their minimal polynomial

    F denotes a field
    V denotes a vector space over F
    L(V) denotes a vector space whose members are linear operators from V to V itself and its field is F, then L(V) is an algebra where multiplication is composition of functions.
    τ, σ, φ denote distinct linear operators contained in L(V)
    m?(x) denotes the minimal polynomial of the linear operator "?"
    ~ denotes the similarity of the left and right operand, e.g., if τ ~ σ, then τ = φσφ-1

    If τ ~ σ are similar linear operators on V, then mτ(x) = mσ(x), i.e., the minimal polynomial is an invariant under similarity of operators.

    I wonder how it's proved.

    Thanks for any help!
  2. jcsd
  3. Jun 20, 2009 #2
    Use dummy variables and factor them wisely, here is a taste

    [itex] \alpha = \alpha\varphi\varphi^{-1}, \alpha\in\mathbb{F},\varphi\in\mathbb{V}[/itex]
  4. Jun 20, 2009 #3
    Last edited by a moderator: Apr 24, 2017
  5. Jun 21, 2009 #4
    Because I am too cool!

    just kidding, it says alpha = alpha*phi*phi^(-1) and alpha in F and phi in V.
  6. Jun 22, 2009 #5
    I still don't understand.
    For example, suppose mτ(x) = s1+s2τ+s3τ2 = 0 and τ = φσφ-1,

    then mτ(x) = s1+s2φσφ-1+s3φ2σ2-1)2 =0,

    multiply (φ2)-1 on the left and φ2 on the right of the both sides of mτ(x)

    we will get s1+s2φ-1σφ+s3σ2 = 0,
    this doesn't equal to s1+s2σ+s3σ2 = 0, does it?

    Ah, yes, now I know where I made a mistake, τ2 ≠φ2σ2-1)2

    since τ2 = (φσφ-1)2 = φσφ-1φσφ-1
    = φσ2φ-1,

    then multiply φ-1 on the left and φ on the right of of the both sides of the equation mτ(x) = s1+s2φσφ-1+s3φσ2φ-1=0

    we will get s1+s2σ+s3σ2 = 0

    The property that τ2 = φσ2φ-1 I've explored before reaching this theorem, it seems I need more practice.

    Thanks for reply, trambolin.

    There are some other questions I posted in "Linear & Abstract Algebra" with no replies yet, maybe you can help me again, thanks a lot!

    (This is not my mother language, please forgive the grammar mistakes I've made)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook