For automorphism T, prove identity map is in span of ${T,T^2,T^3, ...}$ Let $V$ be a vector space over $F$, $T$ be an automorphism with an inverse $S$ s.t. $ST=1_v$. Can I say that $1_v$ is in the span of $\{T^i\;|\; i=1,2,3 ...\}$? 
I feel this is right but I can find a way to reason it.
 A: In general, no - pick a permutation $\pi$ of $\mathbb{N}$ of infinite order$^*$, look at the vector space $V=\bigoplus_\mathbb{N}F$ with standard basis $\{e_i: \in\mathbb{N}\}$, and consider the automorphism $T$ generated by $e_i\mapsto e_{\pi(i)}$.
But! There is a natural additional hypothesis you can make about your vector space, which will make your statement true. Can you see from this example what that hypothesis should be?

$^*$It's a good exercise to show that such a permutation exists. Here's a hint: partition $\mathbb{N}$ into sets $A_i$, with $\vert A_i\vert=i$, for $i\in\mathbb{N}$. For each $i$, let $\pi_i$ be a permutation of $A_i$ of order $i$ - just a cycle of length $i$. Now, what happens when we combine the $\pi_i$s?
A: In finite dimension your statement is true. Take the minimal polynomial $P_T(x)$ of the application $T$. $x$ does not divide $P_T$ because $T$ is invertible and so it's injective. You have that $P_T(T)=0$, so something like the following holds ($c \neq 0$)
$$a_nT^n+...+a_1T+c=0.$$
From there, dividing by $c$, you can easily derive an expression for the identity in function of $T$.
