About Hom(V,V) and group

Question

I was looking an example to show this The set of all non zero elements of Hom(V,V) is not a group under the composition of linear operators.

Answer 1

You seem to be having trouble coming up with specific examples. A linear operator is an operator $A$ such that $$A(ax+by)=aA(x)+bA(y)$$ For example, on $\mathbb R^2$ the function $$A(x,y) = (x,0)$$ is a linear operator (a nonzero element of $\mathrm{Hom}(V,V)$), as is $$B(x,y)=(0,y)$$ But $$AB(x,y)=(0,0)$$ for all $(x,y)$, so the set of nonzero elements of $\mathrm{Hom}(V,V)$ is not closed under composition, since $A\neq 0$, $B\neq 0$, but $AB=0$.

You could also show that $A$ is not invertible and that would disprove that the set is a group as well.

It's worth saying that if $V$ is one-dimensional then the set of nonzero elements of $\mathrm{Hom}(V,V)$ actually is a group, but it is not a group provided $\dim(V)>1$.