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Let $C\in \text{Mat}_{n\times n}(\mathbb R)$. Then which of the alternatives are correct:

  1. $\operatorname{dim}\langle I,C,C^2,\dots,C^{2n}\rangle$ is at most $2n$
  2. $\operatorname{dim} \langle I,C,C^2,\dots,C^{2n}\rangle$ is at most $n$.
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    @copper.hat Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868). – 2013-09-15

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Following Julian's suggestion:

Cayley Hamilton shows that every matrix satisfies its own characteristic polynomial, in particular, $p(C)=0$ for some polynomial $p$ of degree $n$. Hence $C^n$ can be written as a linear combination of $I,C,...,C^{nāˆ’1}$.

It follows by induction that any higher power of $C$ can be written in terms of $I,C,...,C^{nāˆ’1}$.

Hence both statements are correct, as Marc pointed out, but the second is stronger.