Show that Jordan block is idempotent if it is 0 or 1

Question

An idempotent matrix $M$ is one such that $M^2 = M$. A Jordan block has its eigenvalue $\lambda$ on its diagonal and 1 on the superdiagonal. I figure that in order to ensure that $M^2=M$, it makes sense that $\lambda = \{0, 1\}$ since those are the only numbers that are equal to themselves when they are squared.

With this in mind, I squared Jordan blocks with $\lambda = \{0, 1\}$ and found that they indeed weren't idempotent. I assume this would also hold true for larger matrices, but I'm not sure how to show this in a proof.

Is my logic correct so far? If so, how can I expand it? If not, what am I doing wrong?

Answer 1

If you square a $2 \times 2$ Jordan block: $$ M = \pmatrix{\lambda & 1\cr 0 & \lambda},\ M^2 = \pmatrix{\lambda^2 & 2\lambda\cr 0 & \lambda^2}$$ These are never equal. Similarly for an $n \times n$ Jordan block with $n > 2$, if $M$ has eigenvalue $\lambda$, $M^2$ has $\lambda^2$ on the diagonal and $2 \lambda$ on the first super-diagonal. So Jordan blocks of size $\ge 2$ are never idempotent.

Answer 2

Note that a $n\times n$ Jordan block for eigenvalue $\lambda$ acts as follows on the relevant basis vectors $e_1,\ldots,e_n$: $Me_1=\lambda e_1$ and $Me_k=\lambda e_k+e_{k-1}$ for $2\le k\le n$. We conclude $M^2e_1=\lambda^2e_1$ and so from $M^2=M$ indeed $\lambda^2=\lambda$, i.e., $\lambda=0$ or $\lambda=1$. If $n\ge2$ then $M^2e_2=M(\lambda e_2+e_1)= \lambda^2e_2+2\lambda e_1$ so that from the coefficient of $e_1$ we find $2\lambda=1$. As both $2\cdot 0\ne 1$ and $2\cdot 1\ne 1$, this is not possible. We conclude that idempotency can only occur for $n=1$ (and $\lambda\in\{0,1\}$).