Idempotency and Jordan cells

Question

Which Jordan cells $J(\lambda,k)$ are idempotent? And how can I use that to determine the Jordan canonical form of any square idempotent matrix?

Answer 1

Hint. It seems the following.

You can easily calculate the square of a Jordan cell. And when you will do it you will easily found which Jordan cells are idempotents and which Jordan canonical forms of a square idempotent matrix can be.

Answer 2

Hints:

Idempotent matrices over a field can only have eigenvalues lying in $\{0,1\}$, and every idempotent matrix is diagonalizable.

To confirm what this suggests, take the block and decompose it into $\lambda I+I'$, where $I'$ denotes the matrix with $1$ on the superdiagonal.

Computing $(\lambda I+I')^2=\lambda I+I'$, you will discover that $\lambda^2=\lambda$, so that $\lambda\in \{0,1\}$. You will also need to notice that $(I')^2$ is zero on the places where $I'$ is nonzero.