vector space generated by $\{I,A,A^2,\dots,A^{2n}\}$ $A$ be $n\times n$ matrix then the  dimension of vector space generated by $\{I,A,A^2,\dots,A^{2n}\}$ is atmost $n$ right? as $c_0I+c_1A+\dots+c_nA^n=0$ with some nonzero co efficient(cayley hamilton). so $\{I,\dots A^n\}$ will be linearly depenedent so others element $\{A^{n+1},\dots,A^{2n}\}$ may form atmost $n$ dimensional vector space. am I right?  
 A: Once you can express some power $A^k$ as linear combination of lower powers $A^i$ with $0\leq i<k$, there is no hope finding any higher powers either that are linearly independent of those lower powers. That's proved an easy induction: $A^{m+1}=AA^m$ and if $A^m$ is a linear combination of $A^0,\ldots,A^{k-1}$ then $A^{m+1}$ is a linear combination of $A^1,\ldots,A^k$, but by assumption we can express that final term $A^k$ as  a linear combination of $A^0,\ldots,A^{k-1}$, and this does the job. (This is the linear algebra version of the periodicity phenomenon, the fact that if some power of a group element $g$ gives the identity, then every further power of $g$ will be equal to a lower power of $g$ already seen.) Moreover, the first $k$ for which the above happens satisfies $k\leq n$ where $n$ is the dimension of the vector space acted upon; this follows from the Cayley-Hamilton theorem, but can also be shown by a fairly direct induction on the dimension without using that theorem. So the subspace of the vector space of all $n\times n$ matrices spanned by all the powers of $A$ (no need to stop at $A^{2n}$) has dimension $k\leq n$ and has $A^0=I,A^1,\ldots,A^{k-1}$ as basis.
