Existence of a $T$-invariant subspace $W$ of $V$ that corresponds to the factor of the characteristic polynomial of $T$.

Question

Let $V$ be a finite dimensional vector space.

Let $T$ be a linear operator on $V$.

Also let $f(x)$ be the characteristic polynomial of $T$.

If there is a polynomial $g(x)$ such that $g(x)|f(x)$, is it true that there is a $T$-invariant subspace $W$ of $V$ such that the characteristic polynomial of $T|_W$ is $g(x)$?

Answer 1

It is true if the field is algebraically closed, write $f(x)=\Pi_i(X-a_i)^{n_i}$, you can write $V=\oplus_iV_i$ where $V_i=Ker(T-a_i)^{n_i}$, write $g(x)=\Pi_i(X-a_i)^{m_i}, m_i\leq n_i$, there exists $W_i\subset V_i$ stable by $T$ such that $dim(W_i)=m_i$. To see this, you can find a basis $(e_i^1,...,e_i^{n_i})$ of $V_i$ such that the matrix of the restriction of $T$ to $W_i$ is triangular superior, take $W_i=Vect(e_i^1,..,e_i^{m_i})$. Take $W=\oplus_iW_i$.

Answer 2

Yes. First notice that there is no polynomial $h(x)\not=1$ such that $h(x)\mid f(x)$ and $h(T)$ is invertible, otherwise $f(x)/h(x)$ would be an annihilation polynomial, hence $f(x)$ cannot be characteristic.
Since $g(T)$ is not invertible, one may take $W=\text{ker}~g(T)$. It follows that $g(x)$ annihilates $T$ on $W$, and $f(x)/g(x)|_{x=T}$ maps $V$ to $W$. Suppose the characteristic polynomial of $T$ on $W$ is $\phi(x)$. If $\phi(x)\not=g(x)$, then $f(x)\phi(x)/g(x)$ annihilates $T$ on $V$, contradicting the minimality of $f(x)$. So $\phi(x)=g(x)$. Now we see the characteristic polynomial of $T|_W$ is $g(x)$.