2
$\begingroup$

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?

  • 0
    aha, ok, then its simple!2012-06-18

1 Answers 1

0

Providing an answer suggested in comments.

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 $(XM_2)[\alpha\circ\beta]=\alpha\circ\beta\circ M_2[\alpha\circ\beta]=\alpha\circ M_2[\beta\circ\alpha]\circ\beta=\alpha\circ0\circ\beta=0$.