# Can the span of $L(I,A,A^2,\cdots,A^k,..)$ have more than $n$ elements?

Let $A\in M_n$. Can dimension of subspace $L(I,A,A^2,\ldots,A^k,\ldots)$ of $M_n$ can be bigger than $n$?

Using Cayley Hamilton theorem that every matrices $A^n$ can expressed as linear combination of $(A^{n-1},A^{n-2},\ldots,I)$ and $p(A)=0$,

if say opposite,and you say that it have more than n elements so it must exist some $A^{n+1}$ that you can expressed as linear combination, so that mean that we have $0=a_{n+1}A^{n+1}+a_nA^n + \cdots +a_0I$

since $p(A)=0$, $0=a_{n+1}A^{n+1}+p(A)$, that mean that $a_{n+1}A^{n+1}=0$, so it opposite what we think so, we only can have n elements, is this ok?

• Can you break up your big paragraph into smaller pieces to make it more readable please? – max_zorn Sep 8 '18 at 7:09

An arbitrary element of $\operatorname{span}\{I, A, A^2, \ldots\}$ can be written as $f(A)$, where $f$ is a polynomial (of any degree).

Cayley-Hamilton states that $\chi(A) = 0$, where $\chi$ is the characteristic polynomial.

Dividing $f$ with $\chi$, we obtain unique polynomials $q,r$ such that $f(x) = \chi(x)q(x) + r(x)$ and $\deg r < \deg \chi = n$.

Plugging in $A$ we get $$f(A) = \chi(A)q(A) + r(A) = r(A)$$

Therefore, an arbitrary element of $\operatorname{span}\{I, A, A^2, \ldots\}$ can be written as a linear combination of only $\{I, A, A^2, \ldots, A^{n-1}\}$.

Hence $\operatorname{span}\{I, A, A^2, \ldots\} = \operatorname{span}\{I, A, A^2, \ldots, A^{n-1}\}$, and the latter clearly has dimension $\le n$.

If I understand what you wrote correctly, you are assuming that $0 = a_{n+1}A^{n+1} + p(A)$ where $p$ is the same polynomial as the one in Cayley Hamilton. I don't think this follows from what you have written.

Suppose we have the equality $A^n = \sum_{i=0}^{n-1} a_i A^i$ from Cayley-Hamilton. Then observe $$A^{n+1} = A^n A^1 = \sum_{i=0}^{n-1} a_i A^{i+1} = \sum_{i=1}^{n-1}a_{i-1}A^i + a_{n-1} A^n = \sum_{i=1}^{n-1}a_{i-1}A^i + a_{n-1} \sum_{i=0}^{n-1} a_i A^{i}= \sum_{i=0}^{n-1}b_i A^i$$

so $A^{n+1}$ can be expressed as a linear combination of $I,A,\dots,A^{n-1}$. Its not hard to see how this would work for any higher power.

(Note on terminology though, the span is a vector space, and assuming you're working over an infinite field, the span has infinite cardinality. Its basis can have at most $n$ elements.)