Theorem: Let $ T: V \to V $ be nilpotent. Let $ f \in \mathbb{F[x]}. $ Show that $ f(T) $ is invertible iff $ f(0) \neq 0 $
Attempt:
$ ( \leftarrow ) $ Suppose that $ f(T) $ is invertible. Hence for every $ n \in \mathbb{N} $ $ ( f(T) )^n \neq 0 $. $ f(T) \neq 0 $ hence there exists $ v \in V $ s.t. $ f(T)v \neq 0 $. There exist $ a_0,…,a_k $ s.t. $ f(x) = a_0 + a_1 \cdot x + … + a_k \cdot x^k $. Since $ T $ is nilpotent there exists $ q $ s.t. $ T^q =0 $ and $ T^{q-1} \neq 0 $
Suppose for the sake of contradiction that $ f(0) = 0 $, thus $ a_0 =0 $ and $ f(x) = a_1 \cdot x + … + a_k \cdot x^k $. $ f(T)v = a_1 \cdot T v + … + a_k \cdot T^k v^k $. Operate on the last equation with $ T $ about $ q-2 $ times and we'll get $ T^{q-2}f(T)v = a_1 T^{q-1}v $. [ Now i'm stuck and don't know how to continue ].
$ ( \rightarrow ) $ Suppose $ f(0) \neq 0 $. We'll show that $ f(T) $ is invertible. Suppose for the sake of contradiction that $ f(T) $ is invertible, hence there exists $ n \in \mathbb{N} $ s.t. $ ( f(T) )^n =0 $ [ Now i'm stuck and don't know how to continue].
Can you please help me as to how to prove the above theorem?, I learned about eigenvalues but currently I'm studying a new section about nilpotency. I've seen related answers that involve eigenvalues and characteristic polynomial, but they're theorems are different than mine. Thanks in advance for any help!
Best Answer
Let $n \in \mathbb N$ the smallest natural number such that $T^n = 0$
(->) Suppose that $f(T)$ invertible but $f(0) = 0$. Since $f(0) = 0$ we have $ f(x) = x q(x)$ for some polynomial $q(x)$. In particular $(f(x))^n= x^n(q(x))^n$ so $(f(T))^n = T^n(q(T))^n = 0$ and $f(T)$ is nilpotent so it can't be invertible : a contradiction.
(<-) Suppose that $f(0) \neq 0$ then
$$ f(T) = a_0I + T(a_1+a_2T+...+a_{n-1}T^{n-1})$$ where the second term is nilpotent. Now use the fact that if $A$ is nilpotent then $I+A$ is in invertible to finish the proof.
To see why this is the case, let $k$ be an odd number such that $A^k = 0$ and let $g(x) = 1+x^k$. Then $g$ has a root for $x = -1$ $$ g(x) = (1+x)p(x).$$ for some polynomial $p(x)$. Then $$ g(A) = I+0 = (1+A)p(A)$$ so $(1+A)$ is invertible.