Q:

Kernel of characteristic polynomial of linear transformation

Let $F$ be a Field and $V$ be a Vectorspace with $\dim(V)=n$. Let $f:V \to V$ be a linear transformation with the characteristic polynomial $pa(x)$.
It is linear factorized. Let $pa(x)=g(x)*h(x)$ with the greatest common divisier  $\gcd(g,h)=1$.

a) $V_g=ker(g(f))$ and $V_h=ker(h(f))$. Show $V_g \cap V_h=\{0\}$.
b) Show $\dim(V_g)+\dim(V_h)=n$.
c) $V=V_g+V_h$

A:

I'll prove (a) and (c) using Bezout's identity. (b) follows from (a) and (c). Also, I'm going to use slightly different notation. If $T:V\to W$ is a linear transformation and $x\in V$, then I'm going to write $Tx$ instead of writing $T(x)$.
Let $F$ be a field, let $V$ be an $F$-vector space with $\dim(V)=n$, and let $T:V\to V$ be a linear transformation. Let $p(x)$ be the characteristic polynomial of $T$. We're given that $p(x)=g(x)h(x)$ with $GCD(g,h)=1$. Let $V_g=\ker(g(T))$ and let $V_h=\ker(h(T))$. We want to show the following:
(a) $V_g\cap V_h=\{0\}$.
(b) $\dim(V_g)+\dim(V_h)=n$
(c) $V_g+V_h=V$.
Note: it follows from Bezout's identity that there exist polynomials $a(x)$, $b(x)$ with
$$a(x)g(x)+b(x)h(x)=1.$$
Also note that the Cayley-Hamilton theorem says that $p(T)=0$.
(a) Let $x\in V_g\cap V_h$. Our goal is to show $x=0$. Since
$$a(x)g(x)+b(x)h(x)=1,$$
it follows that
$$a(T)g(T)+b(T)h(T)=I,$$
where $I:V\to V$ is the identity transformation. Hence
$$x=Ix=\left[a(T)g(T)+b(T)h(T)\right]x=a(T)g(T)x+b(T)h(T)x=a(T)0+b(T)0=0.$$
Hence $x=0$.
(c) Let $x\in V$. Our goal is to show that $x=y+z$ for some $y\in V_g$, $z\in V_h$. Since
$$g(x)a(x)+h(x)b(x)=1,$$
it follows that
$$g(T)a(T)+h(T)b(T)=I.$$
Hence
$$x=Ix=\left[g(T)a(T)+h(T)b(T)\right]x=g(T)a(T)x+h(T)b(T)x.$$
Let $y=b(T)x$ and $z=a(T)x$. Then $x=y+z$. Also $y\in V_g$ since
$$g(T)y=g(T)h(T)b(T)x=p(T)b(T)x$$
and $p(T)b(T)x=0$, because $P(T)=0$.
Similarly, $z\in V_h$ since
$$h(T)z=h(T)g(T)a(T)x=p(T)a(T)x$$
and $p(T)a(T)x=0$, because $P(T)=0$.
(b) For any vector space $W$ with subspace $W_1$, $W_2$, we have that
$$\dim(W_1+W_2)+\dim(W_1\cap W_2)=\dim(W_1)+\dim(W_2).$$
It follows that
$$
\begin{align*}
\dim(V_g)+\dim(V_h) &=\dim(V_g+V_h)+\dim(V_g\cap V_h) \\
&=\dim(V)+\dim(\{0\}) \\
&=n+0 \\
&=n.
\end{align*}
$$
Final Comment: Note that to prove (a) we only needed that $GCD(h,g)=1$, and to prove (c) we only needed that $GCD(g,h)=1$ and that $g(x)h(x)=p(x)$ for some polynomial $p(x)$ with $p(T)=0$. So claims (a), (b), (c) would hold whenever we have a polynomial $p(x)$ with $p(T)=0$ and $p(x)=g(x)h(x)$ with for some $g$, $h$ with $GCD(g,h)=1$.

