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This is my first post so excuse me for lack of proper formatting.

I tried to prove it by contradiction thus assuming the minimal polynomial is reducible, the primary decomposition theorem gives us a proper invariant subspace contradicting the given assumption.

So I reduced the problem to the case where minimal is $g(t)^m$ where $g(t)$ is irreducible. I showed that image of $g(t)$ is contained in $\ker(g^m-1 (T))$ but im stuck here and dont know how to proceed or if that even helps.

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Suppose the minimal polynomial factors as $p(t)q(t)$. We then have $p(T)(q(T)(x))=0$ for all $x$ in the overall space $V$, so the image of $q(T)$ is contained in the kernel of $p(T)$.

Now, consider the space $X=\{x: p(T)(x)=0\}$, the kernel of $p(T)$. This kernel is an invariant subspace; $0=T(p(T)(x))=p(T)(Tx)$.
If $X$ is all of $V$, then $p(T)$ is identically zero. This means that the minimal polynomial $pq$ of $T$ must divide $p$, and thus $q$ is a constant.
If $X=\{0\}$, then since the image of $q(T)$ is contained in $X$, $q(T)$ is identically zero. This means that the minimal polynomial $pq$ of $T$ must divide $q$, and thus $p$ is a constant.
By hypothesis, those are the only two options. If $0$ and $V$ are the only invariant subspaces, any factorization of the minimal polynomial must be trivial with one of the factors constant, and the minimal polynomial is therefore irreducible.

That's it. We don't need to worry about the nature of the factorization; we just build a "new" invariant subspace out of it, and show that it can only be one of the trivial subspaces if the factorization is trivial.

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