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

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    $\begingroup$ What is the minimal polynomial of a vector space? $\endgroup$ Commented Mar 27, 2023 at 22:00
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    $\begingroup$ And which vector spaces have only trivial subspaces? $\endgroup$ Commented Mar 27, 2023 at 22:03
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    $\begingroup$ 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. $\endgroup$ Commented Mar 27, 2023 at 22:07
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    $\begingroup$ I edited it again, it is clear enough? $\endgroup$ Commented Mar 28, 2023 at 0:51
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    $\begingroup$ 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 $F[t] / \langle p(t) \rangle$. $\endgroup$ Commented Mar 28, 2023 at 2:26

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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.

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