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How could one describe this subalgebra?

I read that it is given by the powers of $A$ :

${\{A^k:k\in \mathbb N}\}$

is it true? why does it form a subalgebra?

Im trying to understand this proof: On algebra dimensions for an endomorphism

where it is said in the accepted answer that $\mathbb Q[A]$ is generated by the powers of $A$

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    In general, that set will not be closed under addition – 2017-01-28
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    Is $V$ a finite dimensional space? – 2017-01-28
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    I just added context with the edit – 2017-01-28
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    @JonathanBaram, the set you wrote is genrally not even a subspace of the endomorphism algebra. For example, unless $A$ is nilpotent it does not contain zero! – 2017-01-28
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    yes I understand this, that is why I was wondering why in this proof http://math.stackexchange.com/questions/2115199/on-algebra-dimensions-for-an-endomorphism/2115210?noredirect=1#comment4350249_2115210 it is said that $\mathbb Q[A]$ is generated as a vector space by the powers of A – 2017-01-28
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    Well, there is a differnce between the set of powers, which is what you wrote, and the *span* of the set of powers. – 2017-01-28

2 Answers 2

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$\newcommand{\A}{\mathfrak{A}}$$\newcommand{\B}{\mathfrak{B}}$$\DeclareMathOperator{\End}{End}$You are looking for the smallest subalgebra $\A$ of $\End(V)$ containing $A$.

Since $\A$ is a subalgebra, the identity $I \in \A$. Since $A \in \A$, and $\A$ is a subalgebra, we have that $A^{2} = A \cdot A \in \A$, $A^{3} = A^{2} \cdot A \in \A$, etc. Thus all $A^{n}$ are in $\A$. Since $\A$ is a subalgebra, all (finite, of course) linear combinations of the $A^{n}$ are in $\A$.

So $\A$ contains the set $\B$ of all linear combinations of the $A^{n}$. Now $\B$ is a subalgebra of $\End(V)$ (sums and products of linear combinations of the powers of $A^{n}$ are themselves such linear combinations), and contains $A$. By the minimality of $\A$, we have $\A = \B$.

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Assume $V$ is a vector space over $\mathbb{F}$. Given $A \in \operatorname{End}(V)$, the algebra $\mathbb{F}[A]$ is defined by

$$ \mathbb{F}[A] := \operatorname{span} \{p(A) \, | \, p \in \mathbb{F}[X] \}. $$

More explicitly, the algebra $\mathbb{F}[A]$ consists of all the operators which are polynomials in $A$ so have the form

$$ a_0 \cdot I + a_1 A + \dots + a_n A^n $$

for some $n \in \mathbb{N}_0$ and $a_i \in \mathbb{F}$. This is a vector space that is spanned by $\{ I = A^0, A, A^2, \dots, \}$ and a commutative algebra (where the multiplication is given by composition of operators).