Subalgebra generated by Rotations Let $\mathcal{L}$ be the algebra of the endomorphisms on a finite vectorial space $V$, and let $\mathcal{R}$ be the subgroup of rotations. 
My first question is:
what is the subalgebra generated by $\mathcal{R}$ when $V = \mathbb{R}^n$? Is it $\mathcal{L}$ itself?
Maybe the Burnside theorem could help (see below).  
I meet this problem in the following context. Given $H$ a subalgebra of $\mathcal{L}$, let $\mathcal{C}(H) := \{ x \in \mathcal{L} \ : \ \forall \ h \in H \ \ xh=hx \} $ be the centralizer of $H$. Note that if $S \subseteq \mathcal{L}$, then $ \mathcal{C}(S) = \mathcal{C}(<S>) $ (subalgbra generated by), so it is not restrictive to study only centralizers of subalgebras. It is straightforward to verify that $\mathcal{C}(H)$ is a subalgebra of $\mathcal{L}$. I would like to show in this way that $\mathcal{C}(\mathcal{R})$ is $<Id>$, i.e. by showing that $<\mathcal{R}> = \mathcal{L}$ (and then using $\mathcal{C}(\mathcal{L}) = <Id>$).
Now denote by $\mathcal{C}^{-1}(X):=\{ Y \mbox{ subalgebra of } \mathcal{L} \ : \ \mathcal{C}(Y) = X \} $. You can easily verify that $ \mathcal{C}(<XY>) = \mathcal{C}(XY) = \mathcal{C}(X) \cap \mathcal{C}(Y) $, so that $\mathcal{C}^{-1}(X)$ is closed under product (of subalgebras) composed with closure. It is not difficult to verify even that is closed under sum by a direct proof.
My second (general, vague and indefinite) question is: what can be said about the "inverse centralizer"? Maybe $\mathcal{C}^{-1}(X)$ is a singleton and I can't see it? This would make $\mathcal{C}$ a bijective function between subalgebras, and would give the study a very different point of view. Maybe is it closed under intersection, such that one can pick the minimal element of the inv. center? In this way to verify that $S \subseteq \mathcal{C}(X)$ is just to verify that $\min \mathcal{C}^{-1}(S) \subseteq X$.
I see that this could seem a naive approach, I'm not confident with algebras! Then, do you know a good introduction to universal algebra?
 A: We can answer your first question by doing some explicit computations, identifying $\mathcal L$ with the algebra of matrices $M_n(\mathbb R)$, and identifying $\mathcal R$ with with orthogonal matrices of determinant 1, i.e. with the special orthogonal group $SO(n)$. If $n=2$, then $$\mathcal R=\left\{\pmatrix{\cos\theta&-\sin\theta\\\sin\theta &\cos\theta} : \theta\in\mathbb R\right\},$$ and it is straightforward to check that $\left\{ \pmatrix{x&-y\\y&x} : x,y\in\mathbb R\right\}$ is the subalgebra generated by $\mathcal R$, so in this case $\mathcal R$ does not generate all of $\mathcal L$.
However, for $n>2$ we can show that $\mathcal R$ does generate all of $\mathcal L$. Let $\mathcal A$ be the subalgebra generated by $\mathcal R$. If $n$ is odd, then $\mathcal R$ contains the matrices
$$\pmatrix{1&0&0&\dots&0\\0&1&0&\dots&0\\0&0&1&\dots&0\\&\vdots&&\ddots&\vdots\\0&0&0&\dots&1}, \pmatrix{1&0&0&\dots&0\\0&-1&0&\dots&0\\0&0&-1&\dots&0\\&\vdots&&\ddots&\vdots\\0&0&0&\dots&-1}$$
so $\mathcal A$ contains their sum, the matrix $2e_{11}$, where $e_{ij}$ is the matrix which has a 1 for the entry in the $i$th row and $j$th column and zeros elsewhere. Now, given any $i$ and $j$, there exist even permutations $\sigma, \tau\in\text{Alt}(\{1,\dots,n\})$ such that $\sigma(1)=i$ and $\tau(1)=j$. If we identify $\sigma$ and $\tau$ with the corresponding permutation matrices, then we have $\sigma,\tau \in\mathcal R$, and $\sigma e_{11} \tau^{-1}=e_{ij}$, so $e_{ij}\in\mathcal A$, which proves that $\mathcal A=\mathcal L$. If $n$ is even, then we just modify the argument slightly: $\mathcal R$ then contains the matrices
$$\pmatrix{1&0&0&0&0&\dots&0\\0&1&0&0&0&\dots&0\\0&0&1&0&0&\dots&0\\0&0&0&1&0&\dots&0\\0&0&0&0&1&\dots&0\\&&\vdots&&&\ddots&\vdots\\0&0&0&0&0&\dots&1},
\pmatrix{1&0&0&0&0&\dots&0\\0&-1&0&0&0&\dots&0\\0&0&-1&0&0&\dots&0\\0&0&0&1&0&\dots&0\\0&0&0&0&1&\dots&0\\&&\vdots&&&\ddots&\vdots\\0&0&0&0&0&\dots&1},
\pmatrix{1&0&0&0&0&\dots&0\\0&-1&0&0&0&\dots&0\\0&0&1&0&0&\dots&0\\0&0&0&-1&0&\dots&0\\0&0&0&0&-1&\dots&0\\&&\vdots&&&\ddots&\vdots\\0&0&0&0&0&\dots&-1},
\pmatrix{1&0&0&0&0&\dots&0\\0&1&0&0&0&\dots&0\\0&0&-1&0&0&\dots&0\\0&0&0&-1&0&\dots&0\\0&0&0&0&-1&\dots&0\\&&\vdots&&&\ddots&\vdots\\0&0&0&0&0&\dots&-1}, $$
so $\mathcal A$ contains their sum $4e_{11}$, and the rest of the proof works as before.
