Two linear functions which do not commute by composition? In my lecture  $(L(E), +, \circ)$ was described as a non commutative ring.
I don't understand why it is not commutative. I can't find any counterexample of two linear functions which don't commute.
 A: Matrix multiplication is not commutative. An example can be found here. Since each $n$-by-$n$ matrix induces a linear map $\mathbb{R}^n \to \mathbb{R}^n$, the corresponding linear maps of the matrices form a counterexample.
A: I assume $E$ is a finite dimensional vector space over a field $F$.
Suppose $E$ has dimension $n>1$, with a basis $\{v_1,v_2,\dots,v_n\}$.
Define the linear maps $f$ and $g$ on the basis, by saying
$$
f(v_1)=v_2,\qquad f(v_i)=v_i, i=2,\dots,n
$$
and
$$
g(v_1)=v_1+v_2,\qquad g(v_i)=v_i, i=2,\dots,n
$$
Then
$$
f\circ g(v_1)=f(v_1+v_2)=f(v_1)+f(v_2)=v_2+v_2=2v_2
$$
whereas
$$
g\circ f(v_1)=g(v_2)=v_2
$$
Since $v_2\ne0$, we have $2v_2\ne v_2$.
The only cases in which $L(E)$ is commutative are when $\dim E=0$ or $\dim E=1$.
If $E$ is not finite dimensional, the same applies (provided we work in a set theory with the axiom of choice).
A: Counterexample in the plane, with orthonormal base $(\vec i, \vec j)$:


*

*Rotate the vector $\vec i$ an angle of $\pi/2$, you get $\vec j$, then get the symmetric w.r.t the subspace $span \{\vec i\}$, and you obtain $-\vec j$.

*Now do it the other way around, take first the symmetric of $\vec i$, it's simply $i$, then rotate an angle $\pi/2$, and you obtain $\vec j$.
Since $\vec j \neq -\vec j$, the two operations, rotation by $\pi/2$ and symmetry w.r.t $span \{\vec i\}$, which are both linear, do not commute.
