# Dimension of a span of matrix powers

Let $A \in M^{C}_{n\times n}$ be a square matrix with a minimal polynomial of degree $k$. What would be $\dim(\mathrm{Span}\{I,A,A^{2},A^{3},...\})$? I think it's $k$ but I'm not sure exactly how to prove it.

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Wouldn't it be nice if the span you are interested in were isomorphic to $\mathbb C[X]/(minimal \; polynomial)$ ? Hmmm, let me see, how would we prove that... – Georges Elencwajg Nov 5 '11 at 18:27

## 2 Answers

Suppose that $c_0I_n+c_1A+c_2A^2+\cdots+c_{k-1} A^{k-1}=0$. Then if $f(x)=c_0+c_1x+c_2x^2+\cdots+c_{k-1} x^{k-1}$, we have $f(A)=0$. But the minimal polynomial has degree $k>k-1$. Therefore, $f(x)=0$ and so $c_0=\cdots=c_{k-1}=0$. This shows that the set $\{I_n, A, A^2, \cdots, A^{k-1} \}$ is linearly independent.

Now suppose $A$ has minimal polynomial $g(x)=a_0+a_1x+\cdots+a_kx^k$. This means $A^k = a_k^{-1}(a_0I_n+\cdots+a_{k-1}A^{k-1})$ and in general $A^m = a_k^{-1}(a_0I_n+\cdots+a_{k-1}A^{k-1})A^{m-k}$ for all $m \geq k$. Thus powers of $A$ bigger than $A^{k-1}$ can always be replaced with lower powers. Thus $\{I_n, A, A^2, \cdots, A^{k-1} \}$ spans as well.

Therefore, the dimension of the span of powers of $A$ is exactly the degree of its minimal polynomial.

Edit: To elaborate on the spanning arugment.

Suppose that $A^n = c_0I_n+c_1A+\cdots+c_\ell A^\ell$ with $c_\ell\not=0$ and $\ell$ as small as possible. Then if $\ell \geq k$, we can write $A^\ell=a_k^{-1}(a_0I_n+\cdots+a_{k-1}A^{k-1})A^{\ell-k}$ and so $A^n = c_0I_n+c_1A+\cdots+c_{\ell-1} A^{\ell-1} + c_\ell a_k^{-1}(a_0I_n+\cdots+a_{k-1}A^{k-1})A^{\ell-k}$. We have expressed $A^n$ using $I_n,A,\dots,A^{\ell-1}$ so $\ell$ was not minimal (contradiction). Thus $\ell<k$.

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It cannot be more than $k$, because the minimal polynomial allows us to write every $A^n$ with $n\ge k$ as a linear combination of lower powers. On the other hand, it cannot be less than $k$ because then there would be a non-trivial linear relation among $I, A, A^2, \ldots, A^{k-1}$, which would be a polynomial with $p(A)=0$ of degree $<k$, contradicting the minimality of the minimal polynomial.

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