Let $V$ be a vector space and $T : V \to V$ a linear transformation. We call $V$ $T$-cyclic if $V$ is generated by $\{ T^i v \}_{i \in \mathbb N}$ for some $v \in V$. For a linear transformation $T : V \to V$ denote by $m_T$ its minimal polynomial and by $f_T$ its characteristic polynomial. Then I know that if $V$ is $T$-cyclic with a minimal polynomial of degree $r$ $$ m_T = a_r x^r + \ldots a_1 x + a_0 $$ then $\{ T^0 v, Tv, \ldots, T^{r-1}v \}$ is a basis for $V$, and $T$ has the representation matrix $$ \begin{pmatrix} 0 & 0 & \ldots & & -a_0 \\ 1 & 0 & \ldots & & -a_1 \\ 0 & 1 \\ \vdots \\ 0 & \ldots & & 1 & -a_r \end{pmatrix} $$ and we have $f_T = m_T$.

Now I have a question on the following proof:

Lemma: Let $V$ be $T$-cyclic and $m_T = gh$, then $\dim \mbox{ker}(h(T)) = \mbox{deg}(h)$.

Proof: Let $U := \mbox{range}(h(T))$, then $U = h(T)V$ is also $T$-cyclic, and similarly $V / U$. With the above we have $$ f_T = m_T, \quad f_{T|U} = m_{T|U}, \quad f_{T|(V/U)} = m_{T|(V/U)}. $$ So $$ gh = m_T = f_T = f_{T|U} f_{T|(V/U)} = m_{T|U} m_{T|(V/U)}. $$ Because of $h(T)V = U$, we have $h(T)V/U = 0$ and therefore $m_{T|(V/U)} | h$. Further $g(T)U = g(T)h(T)V = m_T(T)V = 0$ and so $m_{T|U} | g$. Therefore $$ m_{T|(V/U)} = h, \quad m_{T|U} = g. \qquad (*) $$ We have $\dim U = \deg f_{T|U} = \deg m_{T|U} = \deg g = \deg m_T - \deg h$. Therefore $\deg h = \dim V/U$ and with the homomorphism theorem $V/\mbox{ker}(h(T)) \cong \mbox{range}(h(T)) = U$ and so $$ \dim V/U = \dim \mbox{ker} h(T) $$ which shows (a).

On the part which I marked with (*), does this really follows? Guess this just follows up to units, i.e. in $\mathbb R$ for two $g = a_n x^n + \ldots + a_0, h = b_n x^n + \ldots + b_0$ we could very well have $g\dot h = (1/a_n g) \cdot (a_n h)$ and the polynomials on the RHS divide those on the LHS, but they are not equal?


You are right that from the mere assumption that $m_T = gh$ one cannot deduce that $g,h$ are monic, and hence never prove that they are the minimal polynomial of anything (since minimal polynomials must by definition be monic). This can be repaired by simply requiring $g,h$ to be monic in the lemma (if one is, the other will be, since $m_T$ is monic). Or by replacing them by monic polynomials obtained by multiplication by appropriate scalars; this affects neither $\dim(\ker(h[T]))$ nor $\deg(h)$.

Note that you can get $(*)$ without any hypothesis about cyclic modules. If $h$ is any monic divisor of the minimal polynomial $m_T$ of $T$, and $U=h[T](V)$, then the minimal polynomial of $T|_U$ is $m_T/h=g$, and the minimal polynomial of $T_{/U}$ is $h$. Because (for the first part) $p[T]$ vanishes on $U$ is and only if $p[T]\circ h[T]$ vanishes on all of $V$, while $p[T]\circ h[T]=ph[T]$, and (for the second part) $q[T_{/U}]=0$ if and only if $g[T]\circ q[T]=0$.



May assume the field of scalars algebraically closed.

Break into pieces and reduce to the case $g(x)h(x) = x^n$ and $h(x) = x^m$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.