5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.