## When are the characteristic and the minimal polynomials of a linear map equal?

Posted: May 26, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Problem 1. Let $V$ be a vector space over some field $F$ and suppose that $\dim_F V = n.$ Let $T: V \longrightarrow V$ be a linear transformation. Prove that the minimal and the characteristic polynomials of $T$ are equal if and only if there exists some $v \in V$ such that $v,T(v), \cdots , T^{n-1}(v)$ are linearly independent over $F.$

Solution. Let $p(x)$ be the minimal polynomial of $T$.

Proof of $\Longleftarrow$. We cannot have $\deg p(x) < n$ because then, since $p(T) = 0,$ we’d  have $p(T)(v)=0$ which contradicts our assumption that $v, T(v), \cdots , T^{n-1}(v)$ are linearly independent. Thus $\deg p(x)=n.$

First proof of $\Longrightarrow$. For every $v \in V$ let $I_v= \{f(x) \in F[x]: \ f(T)(v) = 0 \}.$ See that $I_v$ is an ideal of $F[x]$ and thus, since $I_v= \langle f_v(x) \rangle,$ for some $f_v(x) \in F[x].$ Clearly $p(x) \in I_v,$ for all $v \in V$ and so $f_v(x) \mid p(x).$ But the number of divisors of $p(x)$ is finite, which means that the number of distinct elements of the set $\{I_v: \ v \in V \}$ is finite. Let $I_{v_1}, \cdots , I_{v_m}$ be those distinct elements and put

$V_i = \{v \in V: \ f_{v_i}(T)(v)=0 \}.$

See that $V_i$ is a subspace of $V,$ for all $1 \leq i \leq k,$ and $V=\bigcup_{i=1}^m V_i.$ Therefore $V=V_k,$ for some $1 \leq k \leq m.$ Thus $f_{v_k}(T)(v)=0,$ for all $v \in V,$ i.e. $f_{v_k}(T)=0.$ That implies $p(x) \mid f_{v_k}(x),$ by the minimality of $p(x),$ and so $f_{v_k}(x)=p(x)$ because we already know that $f_{v_k}(x) \mid p(x)$ and that $p(x)$ is a monic polynomial. Finally, suppose that

$c_0v_k + c_1T(v_k) + \cdots + c_{n-1}T^{n-1}(v_k)=0,$

for some $c_j \in F$, and let $g(x)=c_0 + c_1x + \cdots + c_{n-1}x^{n-1} \in F[x]$. Then $g(T)(v_k)=0$, which gives us

$g(x) \in I_{v_k} = \langle f_{v_k}(x) \rangle = \langle p(x) \rangle,$

that is $p(x) \mid g(x).$ Therefore $g(x)=0$, because $\deg p(x)=n > \deg g(x),$ and hence $c_j = 0,$ for all $j.$

Second proof of $\Longrightarrow$.  Give $V$ a structure of an $F[x]$ module by defining the multiplication by $f(x)v = f(A)(v).$ Then see that $V$ becomes a finitely generated torsion $F[x]$ module and thus, by the fundamental theorem for finitely generated modules over PIDs, there exists some $v \in V$ such that $\text{ann}_{F[x]}(v)=\text{ann}_{F[x]}(V).$ It’s easy now to show that $v, T(v), \cdots , T^{n-1}(v)$ are linearly independent over $F. \Box$

An important result of Problem 1 is the following:

Problem 2. Let $V$ be a vector space over $F$ and $\dim_F V = n.$ Let $T: V \longrightarrow V$ be a linear transformation. If $T^j, \ 0 \leq j \leq n-1,$ are linearly independent over $F,$ then there exists some $v \in V$ such that $T^j(v), \ 0 \leq j \leq n-1,$ are linearly independent over $F.$

Solution. Since the operators $T^j, \ 0 \leq j \leq n-1,$ are linearly independent, the degree of the minimal polynomial of $T$ is $n$ and thus the characteristic and the minimal polynomials of $T$ are equal. Now the result follows by Problem 1. $\Box$