Why $x_{ss}$ is a polynomial in $x$ ? Let $x$ an endomorphism of a finite-dimensional vector space. The Jordan decomposition state that $x = x_{ss}+ x_n$ where $x_n$ is nilpotent, $x_{ss}$ is semi-simple (diagonalisable). Moreover, these elements commute. In the proof I saw, for show that $x_{ss}$ and $x_n$ commutes, they showed than $x_{ss}$ was a polynomial in $x$. If this is true then I understand why $x_{ss}$ and $x_n$ commute but I don't see how to write $x_{ss}$ as a polynomial in $x$. The argument was : take $\lambda_i$ the set of eigenvalue of $x_{ss}$. Then, there is a polynomial which is zero on $\lambda_i$ with multiplicity big enough and such that $p(0) = 0$. I don't see why this implies that $x{ss}$ is a polynomial in $x$ : I'm surely missing something obvious. I tried to compute $p(x) = 0$, but I don't know how to isolate $x_{ss}$. Thanks in advance !
 A: Let $X$ be a linear operator defined on a finite dimensional vector space $V$ (over some algebraically closed field). Let $\lambda_1,\dots,\lambda_k$ be the distinct eigenvalues of $X$ with respective multiplicities $m_1,\dots,m_k$. 
Then $f(t) = (t-\lambda_1)^{m_1}\cdots (t-\lambda_k)^{m_k}$ is the characteristic polynomial of $X$ (and so $f(X)=0$). 
By Jordan, $V$ is a direct sum of $V_i = \mathrm{Ker}\,(X-\lambda_i I)^{m_i}$ ( generalized eigenspaces). Moreover, each of these subspaces is $X$-invariant ($X$ maps each $V_i$ into itself). 
Restricting ourselves to the subspace $V_i$ (which we can do since it's invariant), the characteristic polynomial (for $X$ restricted to $V_i$) is $(t-\lambda_i)^{m_i}$. 
Apply the Chinese remainder theorem to get a polynomial $p(t)$ such that $p(t)$ is congruent to $\lambda_i$ modulo $(t-\lambda_i)^{m_i}$ for each $i=1,\dots,k$ (and also $p(t)$ congruent to $0$ modulo $t$ if $0$ is not an eigenvalue). Such a $p(t)$ exists since all $(t-\lambda_i)^{m_i}$ (and $t$ if $0$ is not an eigenvalue) are relatively prime.
Now $p(t)$ has zero constant term (due to the final congruence or due to the $0$-eigenvalue congruence). 
Notice that since $V_i$ is $X$-invariant, so is $p(X)$. Next, since $p(t)$ is congruent to $\lambda_i$ modulo $(t-\lambda_i)^{m_i}$, we have $p(t)=\lambda_i + g(t)(t-\lambda_i)^{m_i}$ for some $g(t)$. Thus on the subspace $V_i = \mathrm{Ker}\,(X-\lambda_i I)^{m_i}$, $p(X)=\lambda_i I +g(X)(X-\lambda_i I)^{m_i} = \lambda_i I$. In other words, $p(X)$ acts semisimply on each $V_i$, so $p(X)$ acts semisimply on all of $V$.
Finally, let $q(t)=t-p(t)$. Note: $p(t)$ has no constant term so, $q(0)=0-p(0)=0$ and so $q(t)$ has no constant term. By the above discussion, we can see that $q(X)$ acts as $X-P(X)=X-\lambda_i I$ on $V_i$. so $q(X)^{m_i}=0$ on $V_i$ (by the definition of $V_i$). This means that $q(X)$ is nilpotent on all $V_i$'s and so is nilpotent on $V$.
Thus $X=p(X)+q(X)$ where $p(X)$ is semisimple, $q(X)$ is nilpotent, and both $p(t)$ and $q(t)$ are polynomials with no constant term.
Note: This is the Jordan-Chevalley decomposition. This proof can be found in James Humphrey's Lie algebra text on pages 17-18.
