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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?

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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 $g[T]\circ q[T]=0$.

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HINT:

May assume the field of scalars algebraically closed.

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

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