Vtyjkyuk

Let $Ain M_n$. 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 $0=a_n+1A^n+1+a_nA^n + cdots +a_0I$

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

An arbitrary element of $operatornamespanI, A, A^2, ldots$ 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 $operatornamespanI, A, A^2, ldots$ can be written as a linear combination of only $I, A, A^2, ldots, A^n-1$.

Hence $operatornamespanI, A, A^2, ldots = operatornamespanI, A, A^2, ldots, A^n-1$, and the latter clearly has dimension $le n$.

If I understand what you wrote correctly, you are assuming that $0 = a_n+1A^n+1 + p(A)$ 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
$$A^n+1 = A^n A^1 = sum_i=0^n-1 a_i A^i+1 = sum_i=1^n-1a_i-1A^i + a_n-1 A^n = sum_i=1^n-1a_i-1A^i + a_n-1 sum_i=0^n-1 a_i A^i= sum_i=0^n-1b_i A^i $$

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.)