Let $T:V\to V$ be a linear transformation, where $V$ is an infinite-dimensional vector space over a field $F$. Assume that $T(V)=\{T(v):v\in V\}$ is finite-dimensional.

Show that $T$ satisfies a nonzero polynomial over $F$, that is, there exists $a_0,\dots, a_n\in F$, with $a_n\neq 0_F$ such that $$a_0v+a_1T(v)+\dots+a_nT^n(v)=0_V$$ for all $v\in V$.

I am not very sure how to approach this question. Suppose the dimension of $T(V)$ is $n$. I tried considering the set $\{T(v),T^2(v),\dots,T^{n+1}(v)\}$ which has to be linearly dependent thus there exists $a_i$ such that $a_1T(v)+\dots+a_{n+1}T^{n+1}(v)=0$.

This seems to be similar to what the question whats, except that the polynomial is dependent on $v$, while the question wants a polynomial that works for all $v\in V$.

Thanks for any help.


You're almost there.

Let $v_1, \dots, v_n \in V$ such that $T(v_1), \dots, T(v_n)$ is a basis for $T(V)$.

Let $P_i(T)$ be a polynomial in $T$ such that $P_i(T)(v_i)=0$.

Take $P(X)=XP_1(X)\cdots P_n(X)$.

Take $v \in V$. Then $T(v)=a_1T(v_1)+\dots+a_nT(v_n)$ and so $v = a_1 v_1 + \dots + a_n v_n + u$ with $u \in \ker T$.

Therefore, $P(T)(v) = 0$ because $T,P_1(T),\dots, P_n(T)$ commute.

  • $\begingroup$ Nice. How do we get $v=u+a_1v_1+\dots+a_nv_n$ with $u\in\ker T$? $\endgroup$
    – yoyostein
    Nov 30 '16 at 10:28
  • $\begingroup$ Also, is the point of $V$ being an infinite-dimensional space to prevent us from using Cayley-Hamilton? (since I don't see how the infinite-dimensional fact is being used). Thanks! $\endgroup$
    – yoyostein
    Nov 30 '16 at 10:30
  • 1
    $\begingroup$ I think I know the first: $T(v)=a_1(Tv_1)+\dots+a_nT(v_n)$. Use linearity to show $T(v-a_1v_1-\dots-a_nv_n)=0$, so $v-a_1v_1-\dots-a_nv_n=u$ for some $u\in\ker T$. $\endgroup$
    – yoyostein
    Nov 30 '16 at 10:39

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