 Solution

An endomorphism $\alpha$ is a projection onto its image precisely when it is the identity on $\mathrm{im}\alpha$ and kills a complementary subspace. If ${\alpha }^2=\alpha$, thenso $\alpha$ is the identity on its image. Moreover,and $\mathrm{im}\alpha \cap \ker \alpha =\{0\}$. Henceand $\alpha$ is the projection onto its image along its kernel. Conversely, every projection is the identity after one application, so applying it twice has the same effect: ${\alpha }^2=\alpha$.

Statement (i) is false. On a two-dimensional vector space, take the nonzero nilpotent endomorphismThen ${\alpha }^2={\alpha }^3=0$, but ${\alpha }^2\ne \alpha$, so $\alpha$ is not idempotent.