Hoffman "Linear Algebra": why need such a long proof? I'm reading "Linear Algebra" by Kenneth Hoffman and Ray Kunze.
I don't quite understand why there's a long proof in $\S$6.4 Theorem 6.
First the triangular matrix is defined:

An $n\times n$ matrix $A$ is called triangular if $A_{ij}=0$ whenever $i>j$ or if $A_{ij}=0$ whenever $i<j$.

Then defined triangulable:

The linear operator $T$ is called triangulable if there is an ordered basis in which $T$ is represented by a triangular matrix.

Then there's Theorem 5:

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 $F$.

Now it comes Theorem 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 $p = (x - c_1)
\dots (x - c_k)$ where $c_1, \dots , c_k$ are distinct elements of
  $F$.

The proof is: (the (1)(2).. numbers are added by me)

Proof 
(1) 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). 
(2)To prove the converse, let $W$ be the subspace spanned by all of
  the characteristic vectors of $T$, and suppose $W \ne V$.  ....

What I don't understand is (2) -- why we need such a long proof (details are below) here?
Since Theorem 5 already proved that "minimal polynomial factors $p=(x-c_1)^{r_1}\dots(x-c_k)^{r_k}$, $c_i$ distinct $\Rightarrow$ $T$ is triangulable"; 
this part of Theorem 6 is "minimal polynomial factors $p=(x-c_1) \dots(x-c_k)$, $c_i$ distinct $\Rightarrow$ $T$ is triangulable", 
so we just need to let all the $r_i$ be $1$, isn't it?
Proof details excerpted from Hoffman

(1) 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). 
(2)To prove the converse, let $W$ be the
  subspace spanned by all of the characteristic vectors of $T$, and
  suppose $W \ne V$.
(3)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. 
(4)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 $h$.
(5)Now $p = (x-c_j)q$, for some polynomial $q$. 
(6)Also $q- q(c_j) = (x - c_j)h$.
(7)We have $q(T)\alpha - q(c_j)\alpha = h(T)(T - c_jI)\alpha = h(T)\beta$. 
(8)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 $W$.
(9)Therefore, $q(c_j)\alpha$ is in
  $W$.
(10)Since $\alpha$ is not in $W$, we have $q(c_j) = 0$. 
(11)That contradicts the fact that $p$ has distinct roots. QED.

 A: The answer to your confusion is that triangulable does not imply diagonalisable.
But that does not mean one has to give as complicated a proof as Hoffman and Kunze do. I would personally not derive the result from the theorem 5. Also I would generalise the harder direction as

Whenever an operator $T$ is annihilated by a polynomial $P=(X-c_1)\ldots,(X-c_k)$ with all $c_i$ distinct, it is diagonalisable with set of eigenvalues contained in $\{c_1,\ldots,c_k\}$

(it does not matter whether the annihilating polynomial $P$ is minimal or not). The best proof is probably to use the kernel decomposition theorem which here gives you the decomposition of the whole space as direct sum of the subspaces $\def\I{\mathbf I}\ker(T-c_i\I)$, which (in so far as they are nonzero) are the eigenspaces.

But one can even do without that theorem, using just the following 

Lemma. One has $\dim\ker(T_1\circ\cdots\circ T_k)\leq\dim\ker(T_1)+\cdots+\dim\ker(T_k)$ for any composition of linear operators $T_1,\ldots,T_k$ on $V$.

This follows easily from the case of a composition $g\circ f$ of two operators; see this question, noting that $\dim(\ker(g) \cap \operatorname{im}(f))\leq\dim(\ker(g))$.
Now for our result, since $(T-c_1\I)\circ\cdots\circ(T-c_k\I)=0$, its kernel is all of $V$. The lemma says that $\dim(V)$ is at most the sum of the dimensions of the subspaces $V_i=\ker(T-c_i\I)$. But the nonzero spaces among those are the eigenspaces of$~T$, and the sum of eigenspaces for different eigenvalues is always direct. Therefore $\dim(V_1)+\cdots+\dim(V_k)=\dim(V_1\oplus\cdots\oplus V_k)$. If that dimension is at least $\dim(V)$, then it is equal to $\dim(V)$, and $V_1\oplus\cdots\oplus V_k=V$; then $T$ is diagonalisable. 
