# Calculating the Dimension of a Subspace of $C\in Mat_{n\times n}(\mathbb R)$

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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en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem Also, you only have $n+1$ elements in your first list! – wj32 Dec 7 '12 at 7:35
Apologies. Corrected my question. – Sugata Adhya Dec 7 '12 at 7:47
@wj32: didn't get it. Please elaborate. – Sugata Adhya Dec 7 '12 at 7:58
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 $I,C,...,C^{n-1}$. – copper.hat Dec 7 '12 at 8:16
Both statements are correct, but the second is strictly stronger than the first. – Marc van Leeuwen Dec 7 '12 at 8:43

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