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Let $A,B\in\mathcal{M}_n(\mathbb{R})$.

We suppose that there exists $P\in\mathbb{R}[X]$ with $\deg P\geq1$ such that $P(0)=1,P(A)=AB$.

How can I show that $A$ is invertible and that $AB=BA$ ?

I don't really see how to exploit the information given here to get to the answer, especially $P(0)=1$.. All I have tried haven't worked up to now.

Hippalectryon
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    Is the identity $I+AQ(A)=AB$ more suggestive? (By the way, the exercise does not make much sense if $n\ne p$, you might want to check this.) – Did Oct 01 '14 at 20:20
  • @Did That doesn't ring any bell. What is $Q$ here ? – Hippalectryon Oct 01 '14 at 20:21
  • Maybe 12 seconds (typing included) is a period of intensive reflexion which can be extended a little... – Did Oct 01 '14 at 20:22
  • @Did Oh, I thought you were asking if I knew of any identity related to something like $I+AQ(A)=AB$. My bad. And yes sorry, $n=p$ (edited) – Hippalectryon Oct 01 '14 at 20:23

2 Answers2

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Let us put $Q(X)=P(X)-P(0)=P(X)-1$ , then $X$ divides $Q$ and $Q(X)=XT(X)$ for some polynom $T$

Since $P(A)=AB$ we have $Q(A)=AB-I_n$ thus: $$AT(A)=AB-I_n$$ thus $$A(B-T(A))=I_n$$

That gives $A$ invertible and $A^{-1}=B-T(A)$

(it's known if a matrix is right or left invertible then it's invertible)

We have also : $B=T(A)+A^{-1}$ Then $AB=BA=AT(A)+I_n$

(one can remark that $AT(A)=T(A)A$ since $T(A)$ is a pololynom of $A$)

Mohamed
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If $P(X)=a_0+a_1X+a_2X^2+\dots+a_nX^n$, then you have $1=P(0)=a_0$ and $AB=P(A)=a_0+a_1A+\dots+a_nA^n$.

So $AB -1=A(a_1+a_2A+\dots+a_nA^{n-1})=AM$ for some matrix $M$, or $A(B-M)=I_n$ which gives invertibility of $A$.

Since you have also $(B-M)A=I_n$ and $AM=MA$, you get $AB-AM=BA-MA$ which gives $AB=BA$.

Hippalectryon
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Pedro
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