proof of necessary and sufficient condition for diagonalizability

First, suppose that $T$ is diagonalizable. Then $V$ has a basis whose elements $\{v_1,\mathrm{\dots },v_n\}$ are eigenvectors of $T$ associated with the eigenvalues $\{{\lambda }_1,\mathrm{\dots },{\lambda }_n\}$ respectively. For each $i=1,\mathrm{\dots },n$, as $v_i$ is an eigenvector, its annihilator polynomial is $m_{v_i}=X-{\lambda }_i$. As these vectors form a basis of $V$, we have that the minimal polynomial (http://planetmath.org/MinimalPolynomialEndomorphism) of $T$ is $m_T=\mathrm{lcm}(X-{\lambda }_1,\mathrm{\dots },X-{\lambda }_n)$ which is trivially a product of linear factors.

Now, suppose that $m_T=(X-{\lambda }_1)\mathrm{\dots }(X-{\lambda }_p)$ for some $p$.Let $v\in V$. Consider the $T$ - cyclic subspace generated by $v$, . Let $T_v$ be the restriction of $T$ to $Z(v,T)$. Of course, $v$ is a cyclic vector of $Z(v,T_v)$, and then $m_v=m_{T_v}={\chi }_T$. This is really easy to see: the dimension of $Z(v,T)$ is $r+1$, and it’s also the degree of $m_v$. But as $m_v$ divides $m_{T_v}$ (because $m_{T_v}v=0$), and $m_T$ divides ${\chi }_{T_v}$ (Cayley-Hamilton theorem), we have that $m_v$ divides ${\chi }_{T_v}$. As these are two monic polynomials of degree $r+1$ and one divides the other, they are equal. And then by the same reasoning $m_v=m_{T_v}={\chi }_T$.But as $m_v$ divides $m_T$, then as $m_v=m_{T_v}$, we have that $m_{T_v}$ divides $m_T$, and then $m_{T_v}$ has no multiple roots and they all lie in $k$. But then so does ${\chi }_{T_v}$. Suppose that these roots are ${\lambda }_1,\mathrm{\dots },{\lambda }_{r+1}$. Then $Z(v,T)={\oplus }_{{\lambda }_i}{E}_{{\lambda }_i}$, where $E_{{\lambda }_i}$ is the eigenspace associated to ${\lambda }_i$. Then $v$ is a sum of eigenvectors. QED.