If $T$ has irreducible minimal polynomial, can one decompose the space as a direct sum of invariant subspaces without non-trivial invariant subspaces.

Question

$T: W \rightarrow W$ is a linear operator on a finite dimensional vector space $W$. If the minimal polynomial of linear operator $T$ on $W$ is irreducible, is there a way to decompose $W$ into $$W=W_1\oplus W_2\oplus \cdots\oplus W_n$$ where each $W_i$ is a subspace of $W$, and for each $i$, $W_i$ has no non-trivial $T$-invariant subspaces? Here non-trivial means neither $\{0\}$ nor $W_i$.

Answer 1

Yes such a decomposition is always possible. Here are two approaches.

By the structure theorem of finitely generated modules over a PID (here $K[X]$ with $X$ acting as $T$), the space decomposes as a direct sum of modules isomorphic to some $K[X]/P_i$ for non constant monic polynomials $P_1,\ldots,P_k$ where each $P_i$ (with $i<k$) divides $P_{i+1}$. The minimal polynomial (which annihilates the entire module) is the final $P_k$. As it is given that this minimal polynomial is irreducible, so without non-constant proper divisors, all the $P_i$ must equal $P_k$, and this gives your result.

Alternatively, your $K[X]$ module naturally becomes a $K[X]/\mu$ module for the minimal polynomial $\mu$, since $\mu$ acts as $0$ on it. But if $\mu$ is irreducible then $K[X]/\mu$ is a field$~F$, and we are dealing with a vector space over$~F$. As such it has bases (or cardinal $\dim_F(W)=\dim_K(W)/\deg(\mu)$) and each such basis gives a direct sum of $1$-dimensional $F$-subspaces, which gives you a decomposition as a $K$-vector space as described in the question.