1
$\begingroup$

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

  • 0
    What if $A$ is a matrix $ \in \mathbb{Q}^{n \times n}$ ? – 2017-01-26
  • 0
    @user1952009 what's the problem with this? All endomorphisms of a vector space are represented by some $\dim V\times\dim V$ matrix after choosing a basis. – 2017-01-26
  • 0
    @AdamHughes Yes. And the question is weird because it doesn't say "let $A \in \mathbb{Q}^{n \times n}$" – 2017-01-26
  • 1
    @user1952009 it is common to apply a basis-free approach to these problems, the explicit matrix description is not necessary since the minimal and characteristic polynomials are basis independent, as are things like dimension. – 2017-01-26
  • 0
    To the op: I changed your title because this does not have a connection to field extensions as $\Bbb Q[A]$ is not necessarily a field, eg if $A=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$ then $\Bbb Q[A] = \Bbb Q[x]/(x^2)$. – 2017-01-26
  • 0
    @AdamHughes I don't see the point of making things more complicated than what they are. – 2017-01-26
  • 1
    @user1952009 It's more complicated to have to choose a specific basis just to have to solve a problem. Additionally, there are constructions in higher algebra eg. category theory where you want a basis-independent result to show that the answer depends only on the transformation and **not** on a choice you made (i.e. a choice of basis). This gives a cleaner, more general result which is not muddied by whether or not you need a basis to prove it (you don't). – 2017-01-26
  • 0
    @AdamHughes Here the basis free approach is useless. Don't make things more complicated than what they are. – 2017-01-26
  • 0
    @user1952009 No? I mean, the proof goes through without a hitch easily and quickly. It's **less** complicated if anything. Why would we go through the trouble to tack on the extra fact that endomorphisms **happen** to have matrix representations when that's not needed or relevant? – 2017-01-26
  • 0
    @AdamHughes Come on : "let $A \in \mathbb{Q}^{n \times n}$, what is the dimension of $\mathbb{Q}[A]$ as a vector space / algebra?" is not the same as what the OP wrote – 2017-01-26
  • 0
    @user1952009 Yes they are, $\Bbb Q[A]$ is a sub-algebra of the full endomorphism algebra of $V$ generated (as an algebra) by $A$. If you choose a basis, $A$ is represented by a matrix, yes, but that changes nothing on the structure of $\Bbb Q[A]\cong\Bbb Q[x]/(m_A(x))$ which is basis independent, and in particular it doesn't change the fact that the dimension is $\dim_{\Bbb Q} \Bbb Q[A]=\deg m_A(x)$. – 2017-01-26

2 Answers 2

1

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 $$ A^\ell = (-1)^{n-1} \sum_{i=0}^{n-1} a_i A^{\ell-(n-i)} \in \s\{A^j: 0 \le j \le n\}$$

  • 0
    could you detail why: Induction now gives $$ A^k \in \s\{A^i:0 \le i \le n-1\}, \quad \text{all $k \ge n$}. $$ – 2017-01-26
  • 1
    @JonathanBaram Added the induction – 2017-01-26
1

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.