Proving that a unicellular operator is necessarily irreducible? I would like to prove that a unicellular operator must be irreducible. Here we define $A$ as being a unicellular operator if it is both non-derogatory (minimum annihilating polynomial is equal to characteristic polynomial) and is monosemic (has just one eigen value $\lambda$). Likewise we say $A$ is irreducible if there is no non-trivial reducing subspace relative to $A$, eg. There is no non-trivial subspace $M$ such that both $M$ and $\bar{M}$ are invariant under $A$. I attempted several different routes to arrive at some kind of contradiction but haven't gotten anywhere and feel like I've made no progress. For example I thought that maybe assuming otherwise would lead to a contradiction of $\text{dim } Ker(A-\lambda I) = 1$ which is a requirement for $A$ to be non-derogatory, but have no idea how to get there from just assuming there is a non-trivial reducing subspace.

If $V$ is your vector space and it has a decomposition $V=V_1+V_2$ with $V_1$ and $V_2$ subspaces such that $V_1\cap V_2=\\{0\\}$, and such that both $V_1$ and $V_2$ are invariant for $A$, then the restrictions of $A$ to each subspace define operators $A_1:V_1\to V_1$, $A_2:V_2\to V_2$.

If the dimension of $V$ is $n$ you have $(A-\lambda I)^n=0$ but $(A-\lambda I)^{n-1}\
eq 0$. The first equation implies $(A_j-\lambda I)^n=0$ for $j=1,2$, which in turn implies that $(A_j-\lambda I)^{\dim(V_j)}=0$ for $j=1,2$. This in turn implies that $(A-\lambda I)^{\max\\{\dim(V_1),\dim(V_2)\\}}=0$. Because $(A-\lambda I)^{n-1}\
eq 0$, this implies that $V_1=V$ and $V_2=\\{0\\}$ or vice versa.