Proving if $v, T(v)\, ..., T^{k}(v)$ are linearly dependent for every $v$, then $I, T, ..., T^{k}$ are linearly dependent.

Question

Suppose $V$ is a finite-dimensional vector space. Take a linear operator $T \in L(V)$. Now suppose that we know for every $v \in V$, the set of vectors $\{v, T(v)\, ..., T^{k}(v)\}$ is linearly dependent. I want to show that this would imply that the set of linear operators $\{I, T, ..., T^{k}\}$ is also linearly dependent.

Here are my thoughts so far: Define $A_v = \{ p(x) \in F[x]:p(T)(v)=0\}$ where $F$ is the scalar field of $V$. Since $A_v$ is an ideal of $F[x],$ there exists a unique monic polynomial $g_v$, such that $g_v$ generates $A_v$: $$\langle g_v\rangle=A_v.$$ Now define $G=\{g_v(x): v \in V\}$. If we take $q(x)=\operatorname{lcm}(G)$ (that is, if such a $q$ exists), then it would suffice to show that $\deg(q(x)) \leq k$. If that's proven, then $\{I, T, ..., T^{k}\}$ would be linearly dependent.

Edit: This statement actually follows from the Cyclic Decomposition Theorem (Linear Algebra (Ed2), Hoffman, Kunze, p233). As a corollary to this theorem, we have that there exists a vector $\alpha \in V$ such that $g_\alpha$ is the minimal polynomial of $T$ (Again, Hoffman, p237). Now by the hypothesis, for this $\alpha$ we have a polynomial $p(x)$ of degree at most $k$ such that $p(T)(\alpha)=0$. By the definition of $g_\alpha$, $g_\alpha(x)$ divides $p(x)$. We also know that $g_\alpha(x)=m_{T}(x)$. Thus $m_{T}(x)$ divides $p(x)$ and has a degree of at most $k$. Hence, there exists a polynomial of degree less than or equal to $k$ such that its value at $T$ would be zero, which is equivalent to what we are trying to prove. Although this completes the implication, I'm hoping to find a more elementary proof.

Edit 2: Statement of the Cyclic Decomposition Theorem: Let $T$ be a linear operator on a finite-dimensional vector space $V$ and let $W_0$ be a proper $T$-admissible subspace of $V$. There exist non-zero vectors $\alpha_1, ..., \alpha_r$ in $V$ with respective $T$-annihilators $p_1, ..., p_r$ such that

(i) $V=W_0 \bigoplus Z(\alpha_1; T) \bigoplus ... \bigoplus Z(\alpha_r; T)$;

(ii) $p_k$ divides $p_{k-1}$, $k=2, ..., r$

($Z(\alpha_i; T)$ is the cyclic subspace of $\alpha_i$ (smallest $T$-invariant subspace of $V$ including $\alpha_i$)).

Edit 3: By a $T$-admissible $W$, we mean a $T$-invariant subspace such that for every polynomial $f(x) \in F(x)$, if $f(T)\beta$ is in $W$, then there exists $\gamma \in W$ such that $f(T)\beta=f(T)\gamma$.

Statement of the CDT corollary: Let $T$ be a linear operator on a finite-dimensional vector space $V$. There exists a vector $\alpha$ in $V$ such that the $T$-annihilator of $\alpha$ is the minimal polynomial for $T$.

Answer 1

Given $v\in V$,

$T$ restricts to an endomorphism of $K[T]v$.

Let $f_v$ be the minimal polynomial of $T|_{K[T]v}$.

If $z\in K[T]v\cap K[T]w$ then $f_z$ divides both $f_v$ and $f_w$.

So if $\gcd(f_v,f_w)=1$ then $K[T]v\cap K[T]w=\{0\}$ and hence $f_{v+w} = f_vf_w$.

If $f_v=gh$ then $f_{g(T)v} = h$.

From there you should be able to construct a cyclic vector, that is some $u\in V$ such that $f_u$ is the minimal polynomial of $T$.

From a cyclic vector the result is immediate.

Answer 2

Suppose for the sake of a contradiction that $T$'s minimal polynomial $m$, had degree greater than $k$.

In that case, by tracing through the proof of the rational canonical form theorem for $F[t]$ modules, where $F$ is a field, and we consider the module with action induced by $T$, letting $C(m)$ denote the companion matrix corresponding to $T$'s minimal polynomial (so in rational canonical form, the bottom right matrix), there must then be a vector $v \neq 0$, (i.e. choose the standard basis vector corresponding to the first column of $C(m)$, the bottom extreme right matrix, corresponding to $T$'s minimal polynomial), such that $\{v, Tv, ... , T^{k}v \}$ (corresponding to the first $k+1$ columns of $C(m)$, because by definition this is the rational canonical form of the linear transformation $T$, so that the second column corresponds to $Tv$, and so forth) are linearly independent.

This contradicts your hypothesis, since we must have that $\{v, Tv, ... , T^{k}v \}$ is linearly dependent for all $v \neq 0$.

Therefore $T$'s minimal polynomial $m$ must have degree no greater than $k$, hence $I, T, ... , T^{k}$ are linearly dependent.

Update : I have since replaced my previous response with one making use of the rational canonical form theorem. Hopefully, it at least makes sense, although may be not more elementary than the proposed proof in the original post.

Answer 3

This is just spelling out the details of @reuns answer, but in case it is useful to have that somewhere, here goes:

Definition: Let $V$ is a finite-dimensional $F$-vector space and $T\colon V \to V$ a linear map. For for any $v \in V$, let $$ E_v = \left\{p(T)(v): p \in F[t]\right\} \text{ and } I_v = \left\{p(T) \in F[t]: p(T)(v) = 0 \right\} $$ so that $E_v$ is the smallest $T$-stable subspace which contains $v$ and $I_v$ is an the ideal in $F[t]$. Since $F[t]$ is a P.I.D., $I_v$ has a unique monic generator which we denote by $f_v$. If $m_T$ is the minimal polynomial of $T$ then clearly $f_v\mid m_T$ and hence $m_T = \text{l.c.m.}(f_v: v\in V)$, the monic generator of $\bigcap_{v \in V} I_v$.

Lemma Let $T\colon V\to V$ be as above.

$i)$ If $U$ is $T$-stable and $m_U$ is the minimal polynomial of $T_{|U}$, $m_{V/U}$ the minimal polynomial of $\bar{T}\colon V/U\to V/U$ then $\text{l.c.m.}(m_U,m_{V/U}) \!\mid m_T \!\mid m_U.m_{V/U}$.

$ii)$ If $v \in V$ and let $m_v$ be the minimal polynomial of $T_{|E_v}$. Then $f_v = m_v \!\mid m_T$.

$iii)$ If $w \in E_v$ then $w= r(T)(v)$ for a unique $r\in F[t]$, $\deg(r)<\deg(f_v)$ (taking $\deg(0)=-\infty$) and $f_w = f_v/\text{g.c.d.}(f_v,r)$.

$iv)$ If $v_1,v_2 \in V$ and $f_1=f_{v_1}, f_2 = f_{v_2}$ are coprime, then $E_{v_1} \cap E_{v_2} = \{0\}$ so that $f_{v_1+v_2} = f_1.f_2$.

Proof: Since $m(T)=0$ it induces the zero map on any subspace or quotient of $V$, and thus if $U$ is $T$-stable, $m_U$ and $m_{V/U}$ divide $m_T$. But as $m_{V/U}(T)\subseteq U$, it follows that $m_U(T)m_{V/U}(T)=0$, and hence $m_T \!\mid m_U.m_{V/U}$. Part $i)$ follows immediately.

For $ii)$, if $w \in E_v$ we may write $w = p(T)(v)$ for some $p\in F[t]$. Then $f_v(T)(w) = f_v(T)p(T)(v) =p(T)f_v(T)(v)=0$ so that $f_w\!\mid f_v$, and hence $f_v = \text{l.c.m.}(f_w: w\in E_v) = m_v$ and by part $i)$ we have $m_v \!\mid m_T$.

For $iii)$, if $w=p(T)(v)$ then by the division algorithm $p= q.f_v +r$ for a unique $r\in F[t]$ with $\deg(r)<\deg(f_v)$, and $w=r(T)(v)$. If $s=\text{g.c.d.}(f_v,r)$ then we may write $s=af_v+br$ and so $s(T)(v) = b(T)(w)$, so that $f_w(T)s(T)(v) = b(T)f_w(T)(w)=0$ and hence $f_v \!\mid f_w.s$. But since $(f_v/s)(T)(w)= (f_v/s)(T)r(T)(v) = (r/s)(T)f_v(T)(v)=0$ we also have $f_w \!\mid f_v/s$ hence $f_w = f_v/s$ as required.

For $iv)$, note that if $u \in E_{v_1}\cap E_{v_2}$, then $f_u \!\mid \text{g.c.d.}(f_1,f_2)=1$ hence $u=0$. It follows that $p(T)(v_1+v_2)=0$ if and only if $p(T)(v_1) = p(T)(v_2)=0$ so that $\text{l.c.m.}(f_1,f_2) = f_1.f_2 \!\mid p$, and so $f_{v_1+v_2} = f_1.f_2$.

Claim There exists a vector $v\in V$ with $f_v = m_T$.

Proof 1: We use induction on $\dim(V)$. If $\dim(V)=0$ then the claim is trivial. If $\dim(V)>0$ then $\deg(m_T)>0$ and we may write $m_T = g^k.h$ where $g$ is prime and $g\nmid h$. By part $iv)$ of the Lemma, it is enough to find $u,w$ with $f_u =g^k$ and $f_w =h$.

Step 1: Let $K = \ker(g(T)^k)$. Using the notation of part $i)$ of the Lemma, clearly $m_K\!\mid g^k$, and since $g^k.h=m_T$, we have $h(T)(V)\subseteq K$, hence $m_{V/K}\!\mid h$. Since $\text{g.c.d.}(g^k,h)=1$ it follows from part $i)$ of the Lemma that $m_K=g^k$ and $m_{V/K} = h$.

Step 2: Since $m_K= g^k$, for all $w \in K$ we have $f_w \!\mid g^k$ so that $f_w = g^r$ for some $0\leq r \leq k$. But if $m=\max\{s \in \{0,1,\ldots,k\}: \exists w \in K, f_w = g^s\}$ then $m_K=\text{l.c.m.}(f_w: w\in K) = g^m$, and hence there exists a $w \in K$ with $f_w = g^k$.

Step 3: Since $K \neq 0$ (e.g. as $w\neq 0$) $\dim(V/K)<\dim(V)$, hence if $\bar{T}$ is the map induced by $T$ on $V/K$ by induction we may find $u_1+K \in V/K$ with $f_{u+K}= m_{V/K}=h$. Since it is clear that $f_{u_1+K}\!\mid f_{u_1}$, it follows that $h \!\mid f_{u_1}\!\mid m_T$ so that $f_{u_1} =g^s.h$ for some $s$, $0\leq s\leq k$. Thus if we let $u =g(T)^s(u_1)$ then by part $iii)$ of the Lemma, we have $f_{u} = h$.

Proof 2: If you don't like the inductive nature of the above, then you can argue as follows:

Step 1: Let $m_T = \prod_{i=1}^k p_i^{n_i}$ where the $p_i$ are distinct primes, and let $h_i = \prod_{j\neq i} p_j^{n_j}$. Using part $iv)$ of the Lemma (and induction) it is enough to find vectors $w_i$ with $f_{w_i} = p_i^{n_i}$ for each $i$, $1\leq i \leq k$, as then $v=\sum_{i=1}^k w_i$ will have $f_v = \prod_{i=1}^k f_{w_i} = m_T$.

Step 2: Let $K_i = \ker(p_i(T)^{n_i})$. Since $h_i(T)(V)\subseteq K_i$ we have $m_{V/K_i} \!\mid h_i$ and clearly $m_{K_i} \!\mid p_i^{n_i}$, and hence $m_{V/K_i}$ and $m_{K_i}$ are coprime and so by part $i)$ of the Lemma, $m_{K_i} = p_i^{n_i}$ (and $m_{V/K_i} = h_i$).

Step 3: If $w \in K_i$ then $f_w \!\mid m_{K_i} =p_i^{n_i}(T)(w)=0$, hence $f_w = p_i^r$ for some $r$, $0\leq r\leq n_i$. But if $m = \max\{s \in \{0,1,\ldots,n_i\}: \exists w \in K_i, f_w = p_i^s\}$ then $m_{K_i} = \text{l.c.m.}(f_w:w \in W)= p_i^m$. Hence $m=n_i$ and there exists a vector $w_i \in K_i$ with $f_{w_i} = p_i^{n_i}$.