#Minimal polynomial of a linear map is equal to that of its representation
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crystal bobcat
$\mu_{A}(A)=0=M_B(\mu_{A} (\varphi))$
rough barnBOT
Rotor 😑
crystal bobcat
Use this
marsh stratus
$\mu_{A}(A)=0=M_B(\mu_{A} (\varphi))$
Does this use my idea of proving f_0 | mu_A?
crystal bobcat
Yep because this proves that $\mu_{A}(\varphi)=0$ thus $f_0 | \mu_{A}$
marsh stratus
Oh
rough barnBOT
Rotor 😑
crystal bobcat
Considering $f_0$ is the minimal polynomial of $\varphi$
rough barnBOT
Rotor 😑
marsh stratus
Thanks
crystal bobcat
You’re welcome
there is a more complicated way to do it, you can prove easily that $f_0 | \mu_A$ now to prove that both are equal you can prove that $ K[\varphi]$ and $K[A]$ are isomorphic (this uses the hint) so they have the same dimension that would mean that both their minimal polynomials have same degree so necessarily $f_0=\mu_A$
rough barnBOT
Rotor 😑
crystal bobcat
No need to do it this way I’m just saying stuff that comes to mind your method works better
marsh stratus
Yep because this proves that $\mu_{A}(\varphi)=0$ thus $f_0 | \mu_{A}$
Oh wait also I forgot to say we need f_0 and mu_A having the same leading coefficient
Otherwise f_0 | mu_A and mu_A | f_0 does not imply mu_A = f_0
crystal bobcat
No, in an integer ring $(A,+,\times)$, $a|a’$ and $a’|a$ $\implies a=a’$
rough barnBOT
Rotor 😑
crystal bobcat
What I’m saying is that if one polynomial P divides Q and Q divides P then they are both equal necessarily
non zero polynomial
marsh stratus
No, in an integer ring $(A,+,\times)$, $a|a’$ and $a’|a$ $\implies a=a’$
Yes but we're dealing with polynomials
-x | x and x | -x, no?
crystal bobcat
well regardless by definition the minimal polynomial has leading coefficient 1 by definition
crystal bobcat
Yes indeed I was mistaken my bad
if $a|a’$ and $a’|a$ it just means that $aA=a’A$
rough barnBOT
Rotor 😑
crystal bobcat
Not that they are equal
marsh stratus
if $a|a’$ and $a’|a$ it just means that $aA=a’A$
What's A?
marsh stratus
Alright