# On the minimal polynomial of the composition of two endomorphisms

Let $K$ be a field, $V$ a finite-dimensional vector space and $\alpha, \beta \in \mathrm{End}(V)$. Show that there exists a $T \in K[x]$ such that $\textrm{Min}_{\alpha * \beta} \cdot T = x \cdot \textrm{Min}_{\beta * \alpha}$. ($\operatorname{Min}$ denotes the minimal polynomial of the endomorphismus).

Guess I can choose for $T$ the minimal polynomial of $\beta$, but than I have no idea how to prove that $\textrm{Min}_{\alpha * \beta} \cdot T = x \cdot \textrm{Min}_{\beta * \alpha}$ because I don't see what assumptions about $\alpha$ and $\beta$ can I use? Do you have any hints for me?

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The minimal polynomial of $\beta$ has nothing to do with this. Try writing out explicitly the condition that the minimal polynomial of $\alpha \beta$ has to satisfy and the condition that the minimal polynomial of $\beta \alpha$ has to satisfy and see if you can relate the two. – Qiaochu Yuan Jun 18 '12 at 14:19
For $\alpha\beta$: $Min_{\alpha\beta}(\alpha\beta) = 0$ and for every polynomial $P$ with $P(\alpha\beta) = 0$ follows $Min_{\alpha\beta} ~|~ P$. Similarly for $\beta\alpha$, $Min_{\beta\alpha}(\beta\alpha) = 0$ and for every polynomial $P$ with $Q(\beta\alpha) = 0$ follows $Min_{\beta\alpha} ~|~ Q$. Sorry, but i don't see how to relate them? – Stefan Jun 18 '12 at 14:32
Hint: Let $Q(x) := x \cdot \textrm{Min}_{\alpha \beta}(x)$ what can you say about $Q(\beta \alpha)$? – martini Jun 18 '12 at 14:41
that its equal $(\beta\alpha) \cdot \textrm{Min}_{\alpha\beta}(\beta\alpha)$, but thats all, i know that $a\cdot f(ba) = f(ab)\cdot a$ for every polynomial, but to use that fact i have to shift parantheses, $\beta(\alpha \cdot \textrm{Min}_{\alpha\beta}(\beta\alpha)) = (\beta\alpha) \cdot \textrm{Min}_{\alpha\beta}(\beta\alpha)$, but i am not sure if thats possible. – Stefan Jun 18 '12 at 15:00
@Stefan: That's just associativity of matrix multiplication, and distributivity over sums. – Arturo Magidin Jun 18 '12 at 17:06

It helps a lot to start formulating the question better. You are given two endomorphisms $\alpha\circ\beta$ and $\beta\circ\alpha$ of $V$ with respective minimal polynomials $M_1=\mu(\alpha\circ\beta)$ and $M_2=\mu(\beta\circ\alpha)$, and are asked to show that $M_1$ divides $XM_2$, in other words that $(XM_2)[\alpha\circ\beta]=0$.
Letting $M_2=\sum_ic_iX^i$ one has $(XM_2)[\alpha\circ\beta]=\sum_ic_i(\alpha\circ\beta)^{i+1}=\alpha\circ(\sum_ic_i(\beta\circ\alpha)^i)\circ\beta=0$, because $\sum_ic_i(\beta\circ\alpha)^i=M_2[\beta\circ\alpha]=0$.
If you already know that $\beta\circ P[\alpha\circ\beta]=P[\beta\circ\alpha]\circ\beta$ for any polynomial $P$ (which is proved similarly), you can also argue directly $(XM_2)[\alpha\circ\beta]=\alpha\circ\beta\circ M_2[\alpha\circ\beta]=\alpha\circ M_2[\beta\circ\alpha]\circ\beta=\alpha\circ0\circ\beta=0$.