# Prove the next proposition about f,g two endomorphisms that meet $f∘g=g∘f$…

Be f,g two endomorphisms $$f,g∈End_k (E)$$, where dim (E)=n>0 and finite, such that $$f∘g=g∘f$$. Prove that for all polynomial $$p(x)$$, $$g(ker(p(f))) ⊆ ker(p(f))$$.

What I tried to do is: I want to prove $$∀u ∈ g(ker(p(f)))⇒ u∈ ker(p(f))$$. firstly, How I prove it $$f∘g=g∘f⇒p(f)∘g=g∘p(f)$$?. I write: $$p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0, a_i∈ K$$. By definition, $$p(f)(u)=0$$. By induction: 1. $$n=0$$, so $$g∘p(f)(u)=g(p(f)(u))=g(0)=0$$ because g is lineal. Also $$g∘p(f)(u)=p(f)∘g(u)=0⇒g(u)=0⇒g(u)∈ker(p(f))$$. Is it correct? Could you help me please?

Thank you!

• It looks correct, except that you don't need (and don't actually use) induction at all. – lisyarus May 21 at 15:29
• You haven't proved anything. When you say "By definition $p(f)(u)=0$", that is the thesis of the implication that you are trying to prove. Then you use it in the next sentence, but you don't know if it is true yet. In the next sentence the first implication is false in general, unless $p(f)$ is injective. The proof should be organized like this: Assume that $x\in g(ker(p(f)))$. Then $x=g(u)$ for some $u\in ker(p(f))$. Then $p(f)(x)=p(f)(g(u))=g(p(f)(u))=g(0)=0$. Therefore $x\in ker(p(f))$. – logarithm May 21 at 15:53
• Hi! I undestand you, I have explained so bad. My question is now how can I prove that: $f∘g=g∘f⇒p(f)∘g=g∘p(f)$ ??? – Oriol May 21 at 17:17
• You can prove it by induction on the degree of $p$. If $p$ has degree $0$, then $p(f)$ is just a multiple of the identity, which commute with all endomorphisms, and in particular with $g$. Assume that $q(f)\circ g=g\circ q(f)$ for all polynomials $q$ of degree at most $n$. If $p$ has degree $n+1$, then you can write $p(x)=xq(x)+a$, where $q$ has degree $n$ and $a$ is a scalar. Then $p(f)\circ g=(fq(f)+aI)\circ g=(fq(f))\circ g+aI\circ g=f\circ g\circ q(f)+g\circ(aI)=g\circ f\circ q(f)+g\circ(aI)=g\circ (f\circ q(f)+aI)=g\circ p(f)$. – logarithm May 21 at 17:34