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Let $T$ be a linear operator on a finite-dimensional vector space $V$. Suppose that (a) the minimal polynomial for $T$ is a power of an irreducible polynomial; (b) the minimal polynomial is equal to the characteristic polynomial. Show that no non-trivial $T$ -invariant subspace has a complementary $T$ -invariant subspace.

Here's an answer that has been there before. However I tried doing it in this way and I couldn't find a contradiction. Can someone point out?

My attempt:

Since (b) is satisfied so there is a $\alpha \ne 0$ which is a $T$- cyclic vector of $V$.

Assume that there is a $W \ne 0 , V$ such that $W$ is $T$-admissible then there is a subspace $W'$ such that

$$V = W \bigoplus W'$$

Now $V = sp\{\alpha,T(\alpha),T^2(\alpha),\cdots , T^{n-1}(\alpha)\}$

Then $W = sp\{T^i(\alpha),\cdots T^j(\alpha)\}$ for some $i,\cdots j \in \{1,2,\cdots,n-1\}$

Let $T^j(\alpha)$ be of the largest index s.t. sp$\{T^j(\alpha)\} \subset W$ then $T^{j+1}(\alpha) \in W'$

Then as $T$ is $T$ - invariant then $T(T^j(\alpha))= T^{j+1}(\alpha) \in W$ which is not possible as $W$ and $W'$ are independent subspace.

Antimony
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    Just because a vector space is spanned by a set of vectors doesn't mean a subspace is spanned by a subset of those vectors. Just because a vector is not in a subspace doesn't mean it's in a given complement of that subspace. – blargoner Dec 25 '21 at 17:36

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