Prove that if $A$ is a square matrix whose characteristic polynomial is factored in linear polynomials then $A$ is similar to a triangular matrix. How can I prove this?
I only know this:
Let  $V$  be a finite-dimensional vector space over $\Bbb C$  and  $T∈L(V,V)$. Then there exists a basis $B$  for  $V$  such that  $M(T)$  is upper triangular with respect to $B$
 Thank you very much.
 A: Ok, since nobody goes there, I sacrifice myself.
$\textbf{Theorem 1}$. Let $K$ be a field and $V$ be a $K$-vector space of finite dimension. Let $T\in L(V)$. If all the eigenvalues of $T$ are in $K$ (the roots of the polynomial $\det(T-x I_n)$), then there is a basis of $V$ s.t. the matrix associated to $T$  is upper-triangular.
$\textbf{Proof}$. STEP 1. Let $\lambda\in spectrum(T)$ and $x\in V\setminus \{0\}$ s.t. $Tx=\lambda x$. Then we consider a basis of $V$ in the form $(x,x_2,\cdots,x_n)$.  The associated matrix has its entries in $K$ and is in the form 
$A_n=\begin{pmatrix}\lambda&l\\0&B_{n-1}\end{pmatrix}$. Note that $\det(A_n-xI_n)=(\lambda-x)\det(B_{n-1}-xI_{n-1})$, that implies that the set of the eigenvalues of $B_{n-1}$ is $spectrum(T)\setminus \{\lambda\}\subset K$ (the eigenvalues are written with the multiplicities). 
Let $W=span(x_2,\cdots,x_n)$; note that $B_{n-1}$ is the matrix of $u\in W\mapsto T(u)\mapsto \pi(T(u))\in W$, where $\pi$ is the projection onto $W$.
STEP 2. We consider $\mu\in spectrum(B_{n-1})$ and $y\not= 0$ s.t. $B_{n-1}y=\mu y$ and we start the recurrence...  $\square$
WHAT'S THE POINT?
$\textbf{Theorem 2}$. If $spectrum(T)=(\lambda_i)_i$, then $spectrum(T^k)=(\lambda_i^k)_i$ and $spectrum(e^T)=(e^{\lambda_i})_i$.
$\textbf{Proof}$. According to Theorem 1, we may assume that $T$ is triangularized into $A$ upper-triangular with diagonal $(\lambda_i)_i$. In the sequel, use the fact that the diagonal of a sum (resp. a product) of two upper-triangular matrices $U,V$ is the pairwise sum (resp. the pairwise product) of the elements $u_{i,i},v_{i,i}$ of the diagonals.  $\square$
