# Invariant subspaces and minimal polynomial

I wanted to know whether every invariant subspace $U$ of an endomorphism $A$ with minimal polynomial $m_A= \Pi_{i=1}^n p_i$, where the $p_i$ are mutually coprime polynomials, can be written in the form $U_i = \ker(p_i(A))$?

Further, if there is no invariant subspace of $A$, can we then say that $m_A$ is irreducible?

• The first part of your question is unclear – Bertrand R Aug 31 '13 at 17:29
• sorry, edited it – user66906 Aug 31 '13 at 17:32
• The subspace containing only the zero vector seems to be a trivial counter-example. – EuYu Aug 31 '13 at 17:33
• ah and besides this one? – user66906 Aug 31 '13 at 17:34
• Well, one counter-example suffices to answer your question, but the entire vector space is another counter-example, for non-scalar mappings (although just as trivial). – EuYu Aug 31 '13 at 17:36

For the first part of your question the answer is not really because you see that : $$\ker (p_i(A))\oplus \ker (p_j(A))$$ is also a stable subspace.
For the second part if there is no invariant subspace of A (meaning the only invariant subspaces are $\{0\}$ and the whole space $E$) then if you would write $m_A=pq$ with $p\wedge q=1$ then with kernel lemma : $$E=\ker p(A)\oplus \ker q(A)$$ and both sides of the diret sum are invariant subspaces. Then $m_A$ is irreductible.
If $U$ is an invariant subspace of A then there exist $p$ such that $p|m_A$ and $U=\ker p(A)$ : indeed consider ponctual minimal polynomial $\mu^U_x$ of $A_{|U}$. There exist one $x\in U$ such that $\mu_x^U=\mu^U$ where $\mu^U$ is the minimal polynomial of $A_{|U}$. Then with $\mu^U_x=\mu^U | m_A$ and $$U=\cup_{x\in U} \ker\mu_x^U(A_{|U})=\ker\mu^U(A)$$ (The last equality is non trivial but generalization of union of subspaces is a subspace iif one contain all other)