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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.

This is mainly algebra of polynomials rather than linear algebra. In fact:

Any polynomial $P(X)inmathbb R[X]$ can be written in the form

$$P(X)=lambdaleft(prod_{i=1}^l(X-a_i)^{n_i}right)left(prod_{i=1}^m(X^2+b_iX+c_i)^{q_i}right)$$

where $l,m,n_i,q_iinmathbb N$, $lambda,a_i,b_i,c_iinmathbb R$ 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 $prod_{i=1}^0alpha_i=1$ 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 $P(X)inmathbb R[X]$, you know that it has at least one root $zinmathbb C$. 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 $lambdainmathbb R$ is the leading coefficient of $P(X)$, $n_iinmathbb N$ and $z_iinmathbb C$.

• If all the roots of $P(X)$ are real, we're done: all $z_iinmathbb R$.




• If $P(X)$ has no real root, then all the $z_i$ are in $mathbb Cbackslashmathbb R$. Now, you notice that $P(overline{z_i})=overline{P(z_i)}=0$ because the coefficients of $P(X)$ are real, and because $z_inotinmathbb R$, $overline{z_i}neq z_i$. 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 $(X-z_i)(X-overline{z_i})=X^2-2Re(z_i)X+|z_i|^2.$



• Otherwise, isolate the reals from the non reals, so you can rewrite your polynomial (renaming) as $$P(X)=lambdaleft(prod_{i=1}^l(X-a_i)^{n_i}right)left(prod_{i=1}^m(X-z_i)^{p_i}right)$$ where $lambda,a_iinmathbb R$, $l,m,n_i,p_iinmathbb N$ and $z_iinmathbb Cbackslashmathbb R$. Then you can conclude just as above.