Let $V$ be a vector space over $\mathbb{C}$, we consider it as a $\mathbb{C}[x]$-module by choosing a linear map $\varphi:V\rightarrow V$ and for $f(x)\in\mathbb{C}[x]$ and $v\in V$, define: $$ f\cdot v=f(\varphi(v))$$ Suppose we know that $V$ is written as direct sum of cyclic modules as follows: $$V=\mathbb{C}[x]/\langle g_1(x)\rangle\oplus \cdots\oplus \mathbb{C}[x]/\langle g_r(x)\rangle \tag{$*$}$$ where $g_i(x)$'s are the annihilators of the cyclic modules.

My question is: Do $g_1(x)\cdots g_r(x)$ tell us anything about the Jordan blocks of the map $\varphi$? What about the minimal and characteristic polynomials? Furthermore, if we know the Jordan form of $\varphi$, can we get the decomposition as ($*$) directly?

  • $\begingroup$ I guess you really mean $f\cdot v=f(\varphi)(v)$, that is, first compute the polynomial $f$ on $\varphi$, and then apply the result to the vector $v$. $\endgroup$ – Andreas Caranti Mar 26 '19 at 15:24

The characteristic polynomial will be the product of the $g_{i}$.

The minimal polynomial is slightly more complicated. First note that the decomposition (*) is not unique. But it becomes unique if you insist on the $g_{i}$ being monic, and require the divisibility conditions $$ g_{1} \mid g_{2} \mid \dots \mid g_{r}. $$ Then $g_{r}$ is your minimal polynomial.

Coming to Jordan, you can also write (*) (recall, it is not unique) so that each $g_{i}$ is of the form $g_{i}(x) = (x -\lambda_{i})^{e_{i}}$. Then each such $g_{i}$ corresponds to a Jordan block of size $e_{i}$, with eigenvalue $\lambda_{i}$. This should also answer your last question.

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    $\begingroup$ Thanks!........ $\endgroup$ – qinr Mar 27 '19 at 4:59

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