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.)

  • $\begingroup$ 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. $\endgroup$ May 1, 2013 at 12:00
  • $\begingroup$ @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? $\endgroup$ May 1, 2013 at 12:11
  • $\begingroup$ @AvatarOfChronos: last sentence of comment: not the product, but the least common multiple. $\endgroup$ Jan 12, 2015 at 10:34
  • $\begingroup$ Is not the claim false? For example, let $T$ be the identity operator and $V$ the space of $1\times 1$ matrices over some field. Then the characteristic and minimal polynomial is $x - 1$, and $V$ is a non-trivial $T$-invariant subspace that has the complementary $T$-invariant subspace $\{0\}$. @MarcvanLeeuwen $\endgroup$
    – user0
    Aug 7, 2021 at 15:41
  • 1
    $\begingroup$ @user0 In the context of this question, "non-trivial ... subspace" should be taken to exclude both the zero-dimensional subspace and the whole space (it is the "nonzero proper" of the linked question). The zero-dimensional subspace and the whole space always are $T$-invariant and complementary, so it is natural to exclude both here. $\endgroup$ Aug 7, 2021 at 19:23

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 by the way is the operator that $T$ induces in the quotient modulo this subspace); 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, their degrees are $\dim W_1$ respectively $\dim W_2$; in particular they are 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.

  • $\begingroup$ Dear Marc, in the final paragraph, it seems the proof works for any sum $W_1+W_2=V$. (I think minimal polynomial of any sum of subspaces equals least common multiple of minimal polynomials of restrictions.) At first glance, this seems to be a stronger conclusion than being indecomposable. (That is, not being a sum of proper subspaces.) What am I missing? $\endgroup$
    – Arrow
    Apr 12, 2019 at 19:02
  • $\begingroup$ Regarding the proof (in the first paragraph) that restriction of a cyclic operator to an invariant subspace remains cyclic - it seems to require existence of a minimal polynomial. On the other hand, it seems that over a PID, any submodule of a cyclic module is itself cyclic. So no finite dimension or minimal polynomial assumptions are necessary. Is this correct? $\endgroup$
    – Arrow
    Apr 19, 2019 at 18:37

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