On algebra dimensions for an endomorphism Let $A \in \text{End}(V)$ be an endomorphism, and $\mathbb Q[A]$ a subalgebra in $\text{End}(V)$ generated by $A$.
Is $\mathbb Q[A]$ always at most dim$V$-dimensional? How to prove it
 A: Note that $\def\QA{\mathbf Q[A]}\QA$ is generated as a vector space by the powers 
$$ \{ A^k: k \in \mathbf N \} $$
of $A$. Recall that by the Cayley-Hamilton theorem, we have 
$$ \chi_A(A) = 0$$
where $\chi_A$ is the characteristic polynomial of $A$. Writing 
$$ \chi_A(X) = \det(A-X) = (-1)^n X^n + \sum_{i=0}^{n-1} a_i X^i$$
where $n := \dim V$ we have 
$$ 0 = (-1)^n A^n + \sum_{i=0}^{n-1} a_i A^i \iff A^n = (-1)^{n-1} \sum_{i=0}^{n-1} a_i A^i $$
That is, 
$$ A^n \in \def\s{\mathop{\rm span}}\s\{A^i: i = 0,\ldots, n-1\}. $$
Induction now gives (see below)
$$ A^k \in \s\{A^i:0 \le i \le n-1\}, \quad \text{all $k \ge n$}. $$
That is, 
$$ \QA = \s\{A^k: k \in \mathbf N\} = \s\{A^i: 0 \le i \le n -1 \} $$
As $\QA$ therefore has a generating set with $n$ elements, it is at most $n = \dim V$-dimensional.

Addendum: Suppose that $\ell \ge n$ is given and we have
$$ A^k \in \s\{A^i: 0 \le i \le n\} $$
holds true for $k < \ell$. To prove that $A^\ell \in \s\{A^i: i < n\}$, multiply 
$$ A^n = (-1)^{n-1} \sum_{i=0}^{n-1} a_iA^i $$
by $A^{\ell-n}$, giving 
$$ A^\ell = (-1)^{n-1} \sum_{i=0}^{n-1} a_i A^{\ell-(n-i)} $$
As $\ell - (n-i) < \ell$ for each $i$, the summands on the right hand side are in $\s\{A^j: 0 \le j \le n\}$, there fore 
$$ A^\ell = (-1)^{n-1} \sum_{i=0}^{n-1} a_i A^{\ell-(n-i)} \in \s\{A^j: 0 \le j \le n\}$$
A: Recall that the minimal polynomial $m_A(x)$ divides the characteristic polynomial $p_A(x)$ which has degree $\dim V$ and that

$$\Bbb Q[A]\cong\Bbb Q[x]/(m_A(x))$$

and this is a $\deg m_A(x)\le \deg p_A(x)=\dim V$ algebra.
