If a linear operator $T$ has minimal, characteristic polynomial $q^k$ for $q$ irreducible, a $T$-invariant subspace has no $T$-invariant complement.

Question

Let $T$ be a linear transformation on a finite dimensional vector space $V$ over the field $F$. Let $p_T$ be the minimal polynomial of $T$ and $f_T$ be the characteristic polynomial of $T$. If $p_T = f_T = q^k$ for some irreducible $q$ with $k > 1$, show that no nonzero proper $T$-invariant subspace can have a $T$-invariant complement.

My approach so far is the following: Let $W$ be a nonzero $T$-invariant proper subspace of $V$. Suppose that $W$ has a $T$-invariant complement $W'$.

We know that since $p_T = f_T$, $T$ has a cyclic vector $\alpha$ and that $V$ has a basis $\{\alpha, T\alpha, \dotsc, T^{n-1}\alpha\}$. If $\alpha$ is in $W$ then $W' = \{0\}$ and we are done (and vice versa), so suppose $\alpha = w + w'$ where $w \in W$ and $w' \in W'$.

From here, though, I am not sure how to proceed. Is this the best approach? Is there a better one?

If $V=W_1\dotplus W_2$, $\dim W_j<\dim V$, then the characteristic polynomial of $T|_{W_j}$ must be of the form $q^n$ with $n<k$, but then $q^N(T)=0$ for some $N<k$ (the bigger of these two $n$'s). — , Jan 04 '15 at 05:01
The condition $k>1$ is superfluous, since for $k=1$ the statement is vacuously true (there are no nonzero proper $T$-invariant subspaces in that case). — Marc van Leeuwen, Jan 12 '15 at 10:37

score -1 · Accepted Answer · answered Jan 04 '15 at 15:52

-1

As per Christian's observation in a comment on the origin question:

Let $p_W$ and $p_{W'}$ be the minimal polynomials of $W$ and $T$, respectively. Since $p_T = q^k$ with $q$ irreducible ($k = \dim V$), $p_W = q^m$ and $p_{W'} = q^{m'}$.

Now, since both $W$ and $T'$ are $T$-invariant, i.e., $TW \subseteq W$ and $TW' \subseteq W'$, we must have $m < k$ and $m' < k'$ for otherwise $W = 0$ or $W' = 0$, contrary to our assumptions.

Let $n = \max{m,m'}$; $q^n$ annihilates both $W$ and $W'$, so $q^n$ annihilates $W \oplus W' = V$. However, $n < k$, and $q^k$ is the minimal polynomial of $W$, which is a contradiction.

Therefore, if $W$ is a $T$-invariant subspace, then it cannot have a nontrivial $T$-invariant complementary subspace.

answered Jan 04 '15 at 15:52

user169845

746

I don't think you properly interpreted Christains hint, and I don't understand the thrid paragraph (in which $k'$ should probably be $k$). Where exactly do you use that the minimal polynomial is also characteristic polynomial? – Marc van Leeuwen Jan 12 '15 at 10:40
More precisely the strict inequalities $m,m'<k$ are not justified: a the restriction to a subspace can in general very well have the same minimal polynomial as the full operator. Also the remark $k-\dim V$ is wrong; instead $\dim V=\deg(q^k)=k\deg(q)$ since $q^k$ is (also) the characteristic polynomial. – Marc van Leeuwen Mar 19 '15 at 17:24

If a linear operator $T$ has minimal, characteristic polynomial $q^k$ for $q$ irreducible, a $T$-invariant subspace has no $T$-invariant complement.

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