# Showing no non-trivial t-invariant subspace has a t-invariant complement.

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.)

• 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. Commented May 1, 2013 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? Commented May 1, 2013 at 12:11
• @AvatarOfChronos: last sentence of comment: not the product, but the least common multiple. Commented Jan 12, 2015 at 10:34
• 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 Commented Aug 7, 2021 at 15:41
• @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. Commented Aug 7, 2021 at 19:23

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.

• 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? Commented Apr 12, 2019 at 19:02
• 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? Commented Apr 19, 2019 at 18:37

I also thought of a possible solution for this, but it doesn't use condition (a), which makes me think it might be wrong.

Edit: It is wrong, as someone pointed out below. I mistakenly thought that if neither subspace contains a basis vector, their sum also doesn't contain it - which is not true.

Let T be a linear operator over a finite dimensional vector space V which satisfies (b), and let W be a nontrivial T-invariant subspace of V. Since the minimal polynomial is equal to the characteristic polynomial, T has a cyclic vector v. Therefore, any T-invariant subspace W either does not contain v or is equal to the whole space V. Using the basis $$\{v,Tv,...,T^{n-1}v\}$$ for V (n = dimV), we see that if V = W + W' from some W', one of W or W' must contain v, which implies one of them is equal to V, which implies they have a nonzero intersection and thus W' cannot be a complementary subspace for V. Since W' was arbitrary, W cannot have a complementary T-invariant subspace.

• It is false that "one of $W$ or $W'$ must contain $v$." For example, take $V=\mathbb{R}^2$, $T(x,y) = (-y,x)$ (rotation by $90^{\circ}$). We can take $v=(1,0)$. We can also decompose $V$ as $W+W'$ with $W=\{(x,x)\mid x\in\mathbb{R}\}$ and $W'=\{(x,-x)\mid x\in\mathbb{R}\}$, neither of which contains $v$. You know that $v$ can be written as the sum of a vector in $W$ and one in $W'$; but the union of $W$ and $W'$ does not equal $V$, so that assertion is unwarranted. Commented Jul 28, 2023 at 20:07