Let $V$ be a finite-dimensional vector space over field $k$ and $R = \text{End}_k V$. How do I see that any left ideal of $R$ takes on the form $Rr$ for some suitable element $r \in R$?
Since $R$ is a semisimple ring, every left ideal with split like this: $R = L \oplus L'$. In this decomposition, $1 = e + e'$ where $e \in L$ and $e' \in L'$.
Why does $e^2 = e$ and $(e')^2 = e'$?
And why does $L = Re$?
Write $1=e+e'$, with $e\in L$ and $e'\in L'$. Then $$ e=e(e+e')=e^2+ee' $$ Note that $e^2\in L$ and $ee'\in L'$, so we have $e^2=e$ and $ee'=0$.
Now, take $x\in L$; then $x=x1=xe+xe'$; then $xe'\in L'$ forces $xe'=0$ and $x=xe\in Re$.
The key part is $R=L\oplus L'$, so for $x\in R$ there exist unique $l\in L$ and $l'\in L'$ such that $x=l+l'$.