I am reading the proof for the existence of the Jordan normal form.
I know the Fitting lemma.
I am missing one thing though, I mean the proof in my book seems to have this particular thing missing.
I think I am missing this statement.
Suppose $V$ is a finite dimensional nonzero vector space over the field $F$ which is a subfield of $\mathbb{C}$. Suppose $T$ is a linear map from $V$ to $V$, and $T$ has a single characteristic root $s$ in $\mathbb{C}$ which is also in $F$. Then there exists a positive integer $m$ such that
$V = Ker(T-s.e)^m$,
where $e$ is the identity map of $V$.
Is this statement true? I think it is because the text I am reading seems to assume it is true. How do we prove it?
Since $V$ is finite-dimensional, so is $\operatorname{Hom}(V,V)$. Therefore $\{\operatorname{Id},T,T^2,\ldots\}$ is a linearly dependent family of linear operators, meaning that some non-trivial linear combination $\sum_{k=0}^mc_kT^k$ is zero, i.e., some nonzero polynomial in $T$ evaluates to zero. Among all such polynomials, pick one with the lowest degree and call it $f(x)$. By dividing each coefficient of $f$ by the leading one, we may assume that $f(x)$ is monic. This is known as the minimal polynomial of $T$.
Factorise $f$ as $\prod_{i=1}^m(x-\lambda_i)$, where $m=\deg f$. If $T-\lambda_k\operatorname{Id}$ is nonsingular for some $k\in\{1,2,\ldots,m\}$, then for $g(x)=\prod_{i\neq k}(x-\lambda_i)$, we have $g(T)=(T-\lambda_k\operatorname{Id})^{-1}f(T)=0$, which is a contradiction to the minimality of $\deg f$. Therefore $T-\lambda_i\operatorname{Id}$ must be singular for each $i$. However, by assumption, the only characteristic value of $T$ is $s$. Therefore $\lambda_i=s$ for all $i$. Consequently, $f(x)=(x-s)^m$ and $(T-s\operatorname{Id})^m=f(T)=0$.