Proving that $\dim(\mathrm{span}({I_n,A,A^2,...})) \leq n$

Question

Let $A$ be an $n\times n$ matrix. Prove that $\dim(\mathrm{span}({I_n,A,A^2,...})) ≤ n$

I'm at a total loss here... Can someone help me get started?

Answer 1

Let $F=\mathrm{span}({I_n,A,A^2,...})$. It suffices to prove that $(I_n, A,..., A^{n-1})$ is a family with cardinality $n$ that spans $F$.

By Cayley-Hamilton, one can write $A^n=a_{n-1}A^{n-1}+\ldots+a_0I_n$, hence $$A^n\in \text{span}(I_n, A,..., A^{n-1}).$$

More generally, it is proved by induction on $m$ that $\forall m \geq n, A^m \in \operatorname{span}(I_n, A,..., A^{n-1}) $.

Hence $(I_n, A,..., A^{n-1})$ spans $F$, thus $\dim F\leq n$.