# About the invariance of similar linear operators and their minimal polynomial

1. Jun 20, 2009

### sanctifier

About the invariance of similar linear operators and their minimal polynomial

Notations:
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

Question:
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. Jun 20, 2009

### trambolin

Use dummy variables and factor them wisely, here is a taste

$\alpha = \alpha\varphi\varphi^{-1}, \alpha\in\mathbb{F},\varphi\in\mathbb{V}$

3. Jun 20, 2009

### sanctifier

Last edited by a moderator: Apr 24, 2017
4. Jun 21, 2009

### trambolin

Because I am too cool!

just kidding, it says alpha = alpha*phi*phi^(-1) and alpha in F and phi in V.

5. Jun 22, 2009

### sanctifier

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.