Singular endomorphism has a nonzero two-sided annihilator ($TS=ST=0,S\ne 0$)

Let $T$ be a singular endomorphism of a finite-dimensional vector space. Choose $0\ne u\in \ker T$ and a nonzero linear functional $\lambda$ vanishing on $\mathrm{im}T$, possible because that image of a linear map is proper. Then $S(x)=\lambda (x)u$ is a nonzero rank-one endomorphism with $TS=ST=0$.