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Let $T$ be a linear map on a finite dimensional vector space $V$ over an arbitrary field. Show that the following 3 statements are equivalent.

(1) The minimal polynomial of $T$ is a product of distinct irreducibles.

(2) Each $T$-invariant subspace is admissible. (An invariant subspace $W$ is admissible if for any polynomial $f$ and $v \in V$ such that $f(T)v \in W$, then there exists $w \in W$ such that $f(T)w=f(T)v$.)

(3) For each $T$-invariant subspace, there exists a $T$-invariant complement.

The equivalence between (2) and (3) is a result in Hoffman and Kunze p.232. I am trying to show the equivalence between (1) and (3). However I don't know how to establish either direction. The problem is easier in complex case because the minimal polynomial is always a product of linear factors. So in the real case it requires just a little more work. but in general field I don't know how to proceed in either direction, any hint/help is appreciated.

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Here's a succinct answer, assuming you know about primary decomposition. This allows reducing to the case where there is only one irreducible factor $P\in K[X]$ of the minimal polynomial (which therefore is a power $P^m$ of$~P$). The reduction is possible because the primary decomposition is canonical, and the primary decomposition of any $T$-invariant subspace$~W$ is obtained by intersecting $W$ with the primary components$~V_i$ of $V$; then a $T$-invariant subspace $W'$ is a complement of$~W$ if and only if $W'\cap V_i$ is a complement of $W\cap V_i$ in each$~V_i$, and so (3) holds in the whole space if and only if it holds in every $V_i$.

Now For the direction $(3)\implies(1)$ there is a simple argument proving the contrapositive. Assume the multiplicity$~m$ of$~P$ is larger than$~1$; we need to find a $T$-invariant subspace without $T$-invariant complement. I claim $W=\ker P[T]$ is such a subspace. By the multiplicity assumption $W\neq V$, so a hypothetical $T$-invariant complement $W'$ of $W$ will be of nonzero dimension. The endomorphism$~R$ of$~W'$ obtained by restricting $P[T]$ (which exists by $T$-invariance) is injective, as $\ker R=W'\cap \ker P[T]=W'\cap W=\{0\}$. But then $R^m$ is also injective, which contradicts that fact that it is the restriction of $P[T]^m=0$.

For the direction $(1)\implies(3)$ the question has been reduced to the question Irreducible minimal polynomial implies every invariant subspace has an invariant complement, for which see under the link.

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