Let $V$ be a finite-dimensional vector space over the field $F$ and let $T$ be a linear operator on $V$. Then $T$ is triangulable if and only if the minimal polynomial for $T$ is a product of linear polynomials over $F$.
My attempt: Suppose minimal polynomial of $T$ is $m=(x-c_1)^{r_1}\cdots (x-c_s)^{r_s}$. We need to show $\exists B=\{\alpha_1,…,\alpha_n\}$ basis of $V$ such that $[T]_B=\begin{bmatrix} a_{11}& \cdots & a_{1n}\\ & \ddots & \vdots \\ & & a_{nn}\\ \end{bmatrix}$, i.e. $T(\alpha_j)=a_{1j}\alpha_1+…+a_{jj}\alpha_j$, $\forall 1\leq j\leq n$. Claim: $\forall j\in J_n$, $P(j):\exists \alpha_j\in V$ such that $T(\alpha_j)=\sum_{i=1}^ja_{ij}\alpha_i$. Proof: We use strong induction. Base case: $j=1$. Let $W=\{0\}$. Clearly $W$ is invariant under $T$. By lemma 3 section 6.4, $\exists \alpha_1 \in V$ such that $\alpha_1\notin W$ and $(T-a_{11}I)(\alpha_1)\in W$, where $a_{11}\in \{c_1,…,c_s\}$. So $(T-a_{11}I)(\alpha_1)=T(\alpha_1)-a_{11}\alpha_1=0$. Thus $T(\alpha_1)=a_{11}\alpha_1$. Inductive Step: Assume $P(j)$ holds, $\forall 1\leq j\leq k-1$. Let $W_{k-1}=\text{span}(\{\alpha_1,…,\alpha_{k-1}\})$. Let $\alpha\in W_{k-1}$. Then $\alpha =\sum_{i=1}^{k-1}c_i\alpha_i$. Since $T(\alpha_i)\in \text{span}(\{\alpha_1,…,\alpha_i\})\subseteq W_{k-1}$ (inductive hypothesis), we have $T(\alpha)=\sum_{i=1}^{k-1}c_iT(\alpha_i)\in W_{k-1}$. Thus $W_{k-1}$ is invariant under $T$. By lemma 3 section 6.4, $\exists \alpha_k\in V$ such that $\alpha_k\notin W_{k-1}$ and $(T-a_{kk}I)(\alpha_k)\in W_{k-1}$, where $a_{kk}\in \{c_1,…,c_s\}$. So $(T- a_{kk}I)(\alpha_k)=T(\alpha_k)-a_{kk}\alpha_k=\sum_{i=1}^{k-1}a_{ik}\alpha_i$. Thus $T(\alpha_k)= \sum_{i=1}^{k}a_{ik}\alpha_i$. By mathematical induction, $\forall j\in J_n$, $\exists \alpha_j\in V$ such that $T(\alpha_j)=\sum_{i=1}^ja_{ij}\alpha_i$. We show $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$. If $c_1 \alpha_1+…+c_n\alpha_n=0$. Since $\alpha_n\notin \text{span} (\{\alpha_1,…,\alpha_{n-1}\})$, we have $c_n=0$. Similarly $c_i=0$, $\forall i\in J_n$. So $B$ is independent. Since $\dim (V)=n=|B|$, we have $B$ is basis of $V$. Hence $\exists B=\{\alpha_1,…,\alpha_n\}$ basis of $V$ such that $T(\alpha_j)=\sum_{i=1}^ja_{ij}\alpha_i$, $\forall j\in J_n$.
Conversely, suppose $T$ is triangulable. So $\exists B$ basis of $V$ such that $[T]_B=\begin{bmatrix} a_{11}& \cdots & a_{1n}\\ & \ddots & \vdots \\ & & a_{nn}\\ \end{bmatrix}$. Characteristic polynomial of $T$ is $f:F\to F$ such that $f(x)=\det (xI_n-[T]_B)=(x-a_{11})\cdots (x-a_{nn})$, $\forall x\in F$. By Cayley-Hamilton theorem, $m|f$. So $m= (x-a_{11})^{e_1}\cdots (x-a_{nn})^{e_n}$; $0\leq e_i\leq 1$. Our desired result. Is my proof better version of Hoffman’s proof?
Que: Author used the terminology “product of linear polynomials over $F$“. But linear polynomial need not be monic.