Exercise 7, Section 7.1 of Hoffman’s Linear Algebra 
Let $V$ be an $n$-dimensional vector space, and let $T$ be a linear operator on $V$. Suppose that $T$ is diagonalizable.
(a) If $T$ has a cyclic vector, show that $T$ has $n$ distinct characteristic values.
(b) If $T$ has $n$ distinct characteristic values, and if $\{\alpha_1,…,\alpha_n \}$ is a basis of characteristic vectors for $T$, show that $\alpha =\alpha_1+…+\alpha_n$ is a cyclic vector for $T$.

My attempt: (a) Let $m$ be minimal polynomial of $T$. By theorem 6 section 6.4, $T$ is diagonalizable$\iff$$m$ is of form $(x-c_1)\cdots (x-c_k)$, where $c_i\in F$ and $c_i\neq c_j$, if $i\neq j$. Since $T$ has a cyclic vector, we have $\exists \alpha \in V$ such that $Z(\alpha ;T)=V$. Let $p_\alpha$ be $T$-annihilator of $\alpha$. By theorem 1 section 7.1, $\deg (p_\alpha)=\dim Z(\alpha;T)=n$ and $p_\alpha =m$. So $\deg (m)=n$. Thus $m=(x-c_1)\cdots (x-c_n)$, where $c_i\in F$ and $c_i\neq c_j$, if $i\neq j$. Since characteristic and minimal polynomial of $T$ have  same roots, we have $c_1,…,c_n$ are distinct roots of characteristic polynomial of $T$. That is $c_1,…,c_n$ are distinct eigenvalues of $T$. Hence $T$ has $n$ distinct eigenvalues of $T$.
(b) We need to show $V=Z(\alpha;T)=\text{span}(\{T^k(\alpha)\mid k\geq 0\})$. Let $B=\{\alpha, T(\alpha),…,T^{n-1}(\alpha)\}$. Let $\lambda_1,…,\lambda_n$ are distinct eigenvalues of $T$ and $T(\alpha_i)=\lambda_i\alpha_i$, $\forall i\in J_n$. So $T^i(\alpha)=T^i(\alpha_1+…+\alpha_n)=\lambda_1^i\alpha_1+…+\lambda_n^i\alpha_n$. Suppose $c_0\alpha+c_1T(\alpha)+…+c_{n-1}T^{n-1}(\alpha)=0$, for some $c_i\in F$. Then $$\sum_{i=0}^{n-1}c_iT^i(\alpha)= \sum_{i=0}^{n-1}c_i(\lambda_1^i\alpha_1+…+\lambda_n^i\alpha_n)= \sum_{i=0}^{n-1}c_i \lambda_1^i\alpha_1+…+c_i\lambda_n^i\alpha_n =\left(\sum_{i=0}^{n-1}c_i\lambda_1^i\right)\alpha_1+…+ \left(\sum_{i=0}^{n-1}c_i\lambda_n^i\right)\alpha_n=0.$$ Since $\{\alpha_1,…,\alpha_n\}$ is independent, we have $\sum_{i=0}^{n-1}c_i\lambda_j^i=0$, $\forall j\in J_n$. So $f= \sum_{i=0}^{n-1}c_i x^i\in F[x]$ has $n$ roots, namely $\lambda_1,…,\lambda_n$. Which implies $\deg (f)\geq n$ or $f=0$. Either $\deg (f)\leq n-1$ or $f=0$. So $f=0$. That is $c_i=0$, $\forall 0\leq i\leq n-1$. Thus $B$ is independent. In particular, $|B|=n$. By theorem 4 corollary 2 section 2.3, $\dim Z(\alpha ;T)\geq n$. Since $Z(\alpha;T)$ is subspace of $V$, we have $\dim Z(\alpha;T)\leq n$. Thus $\dim Z(\alpha;T)= n=\dim (V)$. Hence $V=Z(\alpha;T)$. Is my proof correct?
Can we show $\text{span} (B)=Z(\alpha;T)$? First I tired to show $T^n(\alpha)=\lambda_1^n\alpha_1+…+\lambda_n^n\alpha_n \in \text{span}(B)$, but can’t find coefficients to make it work.
 A: (a) Since $T$ has a cyclic vector, we have $\exists \alpha \in V$ such that $Z(\alpha ;T)=V$. So $\dim Z(\alpha ;T)=n$. By theorem 1 section 7.1, $B=\{\alpha ,T(\alpha),…,T^{n-1}(\alpha)\}$ is basis of $Z(\alpha ;T)$. Since $T$ is diagonalizable, $\exists B’=\{\alpha_1,…,\alpha_n\}$ basis of $V$ such that $T(\alpha_i)=k_i\alpha_i$, $\forall i\in J_n$.So $\alpha =\sum_{i=1}^na_i\alpha_i$ and $T^j(\alpha)=\sum_{i=1}^na_ik_i^j\alpha_i$, for all $1\leq j\leq n-1$. Change of basis matrix from $B$ to $B’$ is $$P=\begin{bmatrix}a_1 & k_1a_1 & \cdots & k_1^{n-1}a_1 \\ \vdots & \vdots & \vdots & \vdots \\ a_n & k_na_n & \cdots &k_n^{n-1}a_n \end{bmatrix}.$$ By theorem 7 section 2.4, $P$ is invertible. By exercise 6 section 5.4, $\det (P)\neq 0$. Since $\det$ is $n$-linear, we have $\det (P)=(a_1\cdots a_n)\det \begin{bmatrix} 1& k_1 & \cdots & k_1^{n-1} \\ \vdots & \vdots & \vdots & \vdots \\ 1 & k_n & \cdots &k_n^{n-1} \end{bmatrix}$. Here is proof of determinant of Vandermonde matrix. So $\det (P)=(a_1\cdots a_n)\prod_{i\neq j}(a_i-a_j)\neq 0$. Which implies $k_i\neq k_j$, if $i\neq j$. Thus $k_1,…,k_n$ are distinct eigenvalues of $T$. Hence $T$ has $n$ distinct eigenvalues of $T$.
(b) Suppose $B’=\{\alpha_1,…,\alpha_n\}$ is basis of $V$ such that $T(\alpha_i)=k_i\alpha_i$, $\forall i\in J_n$ and $k_i\neq k_j$, if $i\neq j$. Let $\alpha =\alpha_1+…+\alpha_n$. Then $ T^j(\alpha)=\sum_{i=1}^n k_i^j\alpha_i$, for all $1\leq j\leq n-1$. So
$$\begin{bmatrix}\alpha \\ T(\alpha)\\ \vdots \\ T^{n-1}(\alpha)\\ \end{bmatrix}=\begin{bmatrix}1 & 1& \cdots & 1\\ k_1& k_2& \cdots &k_n\\ \vdots & \vdots & \vdots & \vdots \\ k_1^{n-1}& k_2^{n-1}& \cdots &k_n^{n-1}\end{bmatrix} \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n\end{bmatrix}=Q \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n\end{bmatrix}.$$
Since $\det (Q)=\prod_{i\neq j}(k_i-k_j)\neq 0$, we have $Q$ is invertible. By theorem 8 section 2.4, $B=\{\alpha, T(\alpha),…,T^{n-1}(\alpha)\}$ is basis of $V$. Thus $V=\text{span}(B)=Z(\alpha;T)$.
I saw above solution from this solution book.
A: a) Since $T$ is diagonalizable, let $p$ the minimal polinomial of $T$,
$$p(x)=(x-a_1)\dotsb (x-a_k)$$
where $a_1,a_2, \ldots a_k$ are all distinct characteristics value of $T$. Since $T$ has a cyclic vector. Hence $p$ is also the characteristics polynomial for $T$ which has degree $n$. Therefore $k=n$. That is $T$ has $n$ distinct  characteristic value.
b) Let $a_i$ be  the characteristic value for $T$ corresponding to $\alpha_i$. Then $T^k\alpha_i=a_i^k\alpha_i$. Thus for any polinomial $g$, we have:
$$g(T)\alpha=\sum_{i=1}^{n}g(a_i)\alpha_i.$$
Since $a_i$ are all distinct, we can choose:
$$g_i(x)=\frac{(x-a_1)\dotsb (x-a_{i-1})(x-a_{i+1})\dotsb (x-a_n)}{(a_i-a_1)\dotsb (a_i-a_{i-1})(a_i-a_{i+1})\dotsb(a_i-a_n)}$$
Thus we can have $g_i(a_j)=\delta_{ij}$. Therefore $g_i(T)\alpha=\alpha_i$. Hence $Z(\alpha, T)$ contains a basis $\{\alpha_1,\alpha_2\ldots,\alpha_n\}$ of $V$. So $\alpha$ is a cyclic vector for $T$.
