Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 8F (Linear Algebra)

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.

Statement (ii) is false. The condition is necessary for an idempotent, but it is not sufficient. For example,satisfies $(I-\alpha {)}^2=0$, hence $\alpha (I-\alpha {)}^2=0$, whileover, for example, $\mathbb{Q}$.

Statement (iii) is false over an arbitrary field. Expanding the assumed idempotence givesMultiplying this relation on the left and right by $\alpha$ showsThus $2\alpha \beta =0$, so the conclusion would be true in characteristic different from two. In characteristic two, however, take $\alpha =\beta =I$. Then $\alpha$, $\beta$, and $\alpha +\beta =0$ are idempotent, but $\alpha \beta =I\ne 0$.

Statement (iv) is false because $\alpha \beta =0$ need not imply $\beta \alpha =0$. For example, over any field takeBoth maps are idempotent and $\alpha \beta =0$, butConsequently