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

$$T: W \rightarrow W$$ 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 $$W_i$$.

• What is the minimal polynomial of a vector space? Mar 27 at 22:00
• And which vector spaces have only trivial subspaces? Mar 27 at 22:03
• Can you specify any assumptions on the $W_i$ and their subspaces? For instance, you've tagged with "invariant subspace", which would perhaps provide an interesting question relating to the Jordan blocks of an endomorphism on $W$, but you haven't made that explicit. With no clear assumptions on what type of subspaces you're looking at, this is a rather trivial problem, just amounting to a choice of basis. Mar 27 at 22:07
• I edited it again, it is clear enough? Mar 28 at 0:51
• Hint: if $p$ is the irreducible minimal polynomial of $T$, and $F$ is the (original) scalar field, then $W$ also becomes a vector space over the field $F[t] / \langle p(t) \rangle$, and minimal invariant subspaces of $W$ over $F$ correspond to one-dimensional subspaces of $W$ over $F[t] / \langle p(t) \rangle$. Mar 28 at 2:26

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 $$i) divides $$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.