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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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    http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem Also, you only have $n+1$ elements in your first list!2012-12-07
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    Apologies. Corrected my question.2012-12-07
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    @wj32: didn't get it. Please elaborate.2012-12-07
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    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 $I,C,...,C^{n-1}$.2012-12-07
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    Both statements are correct, but the second is strictly stronger than the first.2012-12-07
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    @copper.hat: Here the subspace is generated by {$I,C,...,C^{2n}$} not by {$I,C,...,C^{n}$}.2012-12-07
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    I understand. It follows that any higher power of $C$ can be written in terms of $I,C...,C^n$.2012-12-07
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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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