Prove that $P_{f}(X) = \prod_{i=1}^{l} (X - a_i)^{n_i} \prod_{j=1}^{m} (X^2 + a_jX + b_j)^{q_j} $. 
Let $f$ be a linear map in a real finite dimension vector space $E$.
Prove that the characteristic polynomial of $f$ can be factorized in the form:
$P_{f}(X) = \prod_{i=1}^{l} (X - a_i)^{n_i} \prod_{j=1}^{m} (X^2 + a_jX + b_j)^{q_j} $


I have thought of using proof by induction.
In one dimensional vector space, it is true.
If we suppose it is true for dimension $n$.
For the case of $n +1$, let $g$ be linear map in an $(n+1)$ dimension vector space.
If we wrote the matrix of $g$ in a basis, we'll have:
$P_g(X) = (a_{11} - X) P(X) + R(X) $
$P(X)$ can be factorized since it is the characteristic polynomial of degree $n$.
I do not know how to get the full factorization of the polynomial.
I might be wrong using the induction, another method would be appreciated.
 A: This is mainly algebra of polynomials rather than linear algebra. In fact:

Any polynomial $P(X)\in\mathbb R[X]$ can be written in the form
$$P(X)=\lambda\left(\prod_{i=1}^l(X-a_i)^{n_i}\right)\left(\prod_{i=1}^m(X^2+b_iX+c_i)^{q_i}\right)$$
where $l,m,n_i,q_i\in\mathbb N$, $\lambda,a_i,b_i,c_i\in\mathbb R$ and $b_i^2-4c_i<0$. (notice that $\lambda$ is the leading coefficient of $P(X)$)

For the above to hold, keep in mind the convention $\prod_{i=1}^0\alpha_i=1$ for any $\alpha_i$.
Now for your exercise, you know that the leading coefficient of the characteristic polynomial is $\lambda=1$.
To prove the fact stated above, use the fundamental theorem of algebra, which says that any non constant polynomial over $\mathbb C$ has at least one root. In particular, given a non constant polynomial $P(X)\in\mathbb R[X]$, you know that it has at least one root $z\in\mathbb C$. So $P(X)=(X-z)Q(X)$ with $\deg Q(X)<\deg P(X)$. With this you can prove by induction that $$P(X)=\lambda \prod_{i=1}^n(X-z_i)^{n_i}$$ where $\lambda\in\mathbb R$ is the leading coefficient of $P(X)$, $n_i\in\mathbb N$ and $z_i\in\mathbb C$.

*

*If all the roots of $P(X)$ are real, we're done: all $z_i\in\mathbb R$.


*If $P(X)$ has no real root, then all the $z_i$ are in $\mathbb C\backslash\mathbb R$. Now, you notice that $P(\overline{z_i})=\overline{P(z_i)}=0$ because the coefficients of $P(X)$ are real, and because $z_i\notin\mathbb R$, $\overline{z_i}\neq z_i$. So you can actually write your polynomial in the form $$\prod_{i=1}^m(X-z_i)^{q_i}(X-\overline{z_i})^{q_i}.$$ Notice that $(X-z_i)(X-\overline{z_i})=X^2-2\Re(z_i)X+|z_i|^2.$


*Otherwise, isolate the reals from the non reals, so you can rewrite your polynomial (renaming) as $$P(X)=\lambda\left(\prod_{i=1}^l(X-a_i)^{n_i}\right)\left(\prod_{i=1}^m(X-z_i)^{p_i}\right)$$ where $\lambda,a_i\in\mathbb R$, $l,m,n_i,p_i\in\mathbb N$ and $z_i\in\mathbb C\backslash\mathbb R$. Then you can conclude just as above.
