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I was reading the answer to another question. In the third paragraph it goes:

...this does not yet make the cyclic decomposition unique, but it does make the sequence of annihilators unique, and therefore the associated rational form.

So let's say we have two cyclic decompositions(with no special properties) that have the same sequence of annihilators, is it possible that the subspaces $V$ is decomposed into are different, i.e. wouldn't the annihilators uniquely determine the subspaces? Can you give such an example?

Would the answer to the questions above be different if the decompositions were in rational canonical form?

J. Doe
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Recall that a cyclic factor of a vector space equipped with a linear operator $T$ is a subspace $W$ generated by some vector $v$ and its repeated images $T^k(v)$ by $T$ (which also has a $T$-stable complementary subspace so that calling it a "factor" is in order), and that the annihilator of $W$ is a minimal degree monic polynomial $P$ such that $P[T](v)=0$ (which implies that $P[T]$ vanishes on all of $W$). The simplest case of a cyclic subspace is the subspace generated by an eigenvector, and its annihilator is $X-\lambda$ where $\lambda$ is the eigenvalue.

So a special case of your question is: can a space have different decompositions as a sum of such $1$-dimensional subspaces, with the same (multi-)set of annihilators $X-\lambda$? The answer is that it clearly can, provided some of the annihilators are the same. Taking $T=\lambda I$, every nonzero vector spans a cyclic subspace with annihilator $X-\lambda$, and every basis of your space$~V$ defines a decomposition into $n=\dim V$ cyclic factors, all with the same annihilator. So there you have infinitely many different such decompositions.

Marc van Leeuwen
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