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My professor stated the following:

For $A\in \mathbb R^{n\times n}$ and $k\geq0$ it can be shown that the $k$th power of $A$, $A^k$, is a linear combination of $I,A,\ldots,A^{n-1}$. A proof is not difficult using the jordan form of $A$.

I have tried to solve this problem to no avail. Recall that the jordan form of $A$ is given as: $A = SJS^{-1}$ where $J$ is the jordan normal form.

Any help would be greatly appreciated.

PS this is not a homework question, the professor just casually said that this holds but I jut want to understand why this holds.

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  • $\begingroup$ See en.wikipedia.org/wiki/…. $\endgroup$ – lhf Sep 29 '15 at 18:43
  • $\begingroup$ Jordan form gives a nice way to show it, but I think Caley-Hamilton is enough. In particular this result kind of shows why the Krylov subspace methods for solving equation systems (i.e. multiplying with $A^{-1}$) can work by performing a set of operations involving iterative multiplication with $A^k$ for positive powers $k$. $\endgroup$ – mathreadler Sep 30 '15 at 10:45
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Here is an outline:

If $B$ is a Jordan block of order $m$ and with $\lambda$ in the diagonal, then $(B-\lambda I)^m=0$.

Take the product of all $(X-\lambda)^m$ corresponding to the Jordan blocks of $A$ and get a polynomial $p(X)$ of degree $n$ such that $p(J)=0$.

Finally, note that $A^k = SJ^kS^{-1}$ and so $p(A)=p(J)=0$.

This is the Cayley–Hamilton theorem, and is equivalent to what you want to prove.

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  • $\begingroup$ Could you explain why it is equivalent? I really don't see that. $\endgroup$ – C_Eng Oct 4 '15 at 8:44

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