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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

An essential hypothesis is (b), implying that no nonzero polynomial of degree less than the dimension of the vector space annihilates$~T$ (as such a degree is incompatible with being a multiple of the characteristic polynomial). Let me call this condition, which has many equivalent statements, that $T$ is cyclic (actually it is the $K[X]$-module defined by$~T$ that is cyclic, but I don't want to mention $K[X]$-modules here). One basic fact is that the restriction of a cyclic operator$~T$ to any $T$-invariant subspace is still cyclic (as is the operator that $T$ induces in the quotient); let me prove that first.

If the restriction of$~T$ to a $T$-invariant subspace of dimension$~d$ were annihilated by a polynomial$~P[T]$ with $\deg(P)<d$, then the image$~W$ of$~P[T]$ would be a $T$-invariant subspace of dimension at most$~\dim V-d$ (by rank-nullity), and annihilated by some$~Q$ with $\deg(Q)\leq\dim W$ (for instance the characteristic polynomial of $T|_W$). But then $QP$ annihilates$~T$ (as $P[T]$ maps $V$ into $W$ which is contained in the kernel of $Q[T]$), and $\deg(QP)<\dim V$, contradicting the hypothesis that $T$ is cyclic.

Now for the actual question. Let $P$ be the irreducible polynomial of point (a), which I may suppose monic, and $P^k$ the minimal polynomial of$~T$. Suppose for a contradiction that $V$ decomposes as a direct sum $W_1\oplus W_2$ of two proper $T$-invariant subspaces. The minimal polynomials of the restrictions of$~T$ to the summands both divide the minimal polynomial $P^k$, and since by the above these restrictions are cyclic, they must be proper monic divisors of$~P^k$. But from the irreducibility of$~P$ this implies they are of the form$~P^l$ with $l<k$. But then their least common multiple, which gives the minimal polynomial of$~T$, cannot be $P^k$, a contradiction.

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