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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 complement 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
@AvatarOfChronos: last sentence of comment: not the product, but the least common multiple. –  Marc van Leeuwen Jan 12 at 10:34

1 Answer 1

Let $P$ be the irreducible polynomial of point (a), and $P^k$ the minimal polynomial. The minimal polynomial of the restriction of$~T$ to any $T$-invariant subspace divides $P^k$. Given that the minimal polynomial of$~T$ is equal to its characteristic polynomial one has $\deg(P^k)=\dim V$. Therefore the minimal polynomial of the restriction of$~T$ to any $T$-invariant proper subspace is a proper divisor of $P^k$, as its degree cannot exceed the dimension of that subspace. Now if $V$ had a decomposition as direct sum of $T$-invariant proper subpaces, this would imply that $P^k$ is the least common multiple of two of its proper divisors, but this is not the case.

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