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.
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.