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The question is from Hoffman and Kunze

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

I know from a,b that T is not diagonalizable; possible irrelevant.

I know that every T-admissible subspace has a complementary subspace which is also invariant under T. So I basically want to show that W={0} and its compliment are the only T-admissible subspaces. Not sure how to do this as T-admissible requires T-invariant.

Can somebody point me in the right direction for how to solve this problem?

(preferable without posting a solution.)

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You don't know that $T$ is not diagonalizable: perhaps the characteristic polynomial is irreducible. Note that $T$ induces a linear mapping on any $T$-invariant subspace, and think about the minimal polynomial of $T$ on the subspace, and on any complement. –  Chris Godsil May 1 '13 at 12:00
    
@ChrisGodsil : Good point about T being diagonalizable, I misunderstood "irreducible" to imply non-linear which we have a theorem about. For T-invariant subspace the minimal polynomial of T restricted to that subspace must divide the minimal polynomial of T on the whole space. The product of a minimal polynomial restricted T-invariant subspace and its compliment should be the minimal polynomial of the space. right? –  AvatarOfChronos May 1 '13 at 12:11

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