Theroem 6: Let V be a finite-dimensional vector space over the field F and let T be a linear operator on V. Then T is diagonalizable if and only if the minimal polynomial for T has the form $$ p = (x-c_1)...(x-c_n) $$ where $c_1,...c_n$ are distinct elements of $\mathbb{F}$.
Proof from Hoffman, Kunze:
We have noted earlier that, if $T$ is diagonalizable, its minimal polynomial is a product of distinct linear factors (see the discussion prior to Example 4).
To prove the converse, let $W$ be the subspace spanned by all of the characteristic vectors of $T$, and suppose $W \ne V$.
By the lemma used in the proof of Theorem 5, there is a vector $\alpha$ not in $W$ and a characteristic value $c_j$ of $T$ such that the vector $\beta= (T - c_jI)\alpha$ lies in W.
Since $\beta$ is in $W$, $\beta = \beta_1+\dots\beta_k$ where $T\beta_i = c_i\beta_i$, $1\le i\le k$, and therefore the vector $h(T)\beta = h(c_1)\beta_1+\dots+h(c_k)\beta_k$ is in $W$, for every polynomial $h$.
Now $p = (x-c_j)q$, for some polynomial $q$.
Also $q- q(c_j) = (x - c_j)h$.
We have $q(T)\alpha - q(c_j)\alpha = h(T)(T - c_jI)\alpha = h(T)\beta$.
But $h(T)\beta$ is in $W$ and, since $0 = p(T)\alpha = (T - c_jI)q(T)\alpha$, the vector $q(T)\alpha$ is in $W$.
Therefore, $q(c_j)\alpha$ is in $W$.
Since $\alpha$ is not in $W$, we have $q(c_j) = 0$.
That contradicts the fact that $p$ has distinct roots.
Can you explain why it is given that $q(x) - q(c_j) = (x-c_j)h(x)$? I don't understand this part so I don't quite get the remaining proof of the theorem.
Note that $c_j$ is a root of $q(x)-q(c_j)$. Hence $q(x)-q(c_j)=(x-c_j)h(x)$ for some polynomial $h$.