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$A$ 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?

Myshkin
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    Yes.${}{}{}{}{}$ – Gerry Myerson May 25 '13 at 07:02
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    (1) There's an $,2n,$ power in the title, where it seems to be just $,n,$ ; (2) Even if itwas $,n,$, why would the dimension of that space be $,n,$ ? It could be less, it could be more, as this space is in $,M_n(\Bbb F),$ , which is an $,\Bbb F$-vector spacve of dimension $,n^2,$ ...? Or are you relying on Cayley-Hamilton from the beginning? – DonAntonio May 25 '13 at 07:19
  • @GerryMyerson I agree with DonAntonio, because the nullity of the vector space is 1, which means the rank is $(n+1)-1=n$, right? Or am I missing something? – rurouniwallace May 25 '13 at 07:53
  • The dimension is the degree of minimal polynomial of $A$. – Sungjin Kim May 25 '13 at 08:18
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    @Zetta, "nullity" is a property of matrices, not vector spaces, so I don't know what you mean. The rank is at most $n$. It could be less, it can't be more. – Gerry Myerson May 25 '13 at 08:51
  • @GerryMyerson I'm talking about the nullity of the matrix formed by the basis of the vector space. – rurouniwallace May 25 '13 at 08:52
  • @Zetta, we have a vector space whose elements are matrices. The basis of that vector space is a set of matrices. What matrix are you forming from those matrices? – Gerry Myerson May 26 '13 at 07:43

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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.

Marc van Leeuwen
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