$A$ is unipotent iff all of its eigenvalues are 1

Question

Note: A is an element of the ring of n-by-n matrices with complex entries.

I know how to prove that $A$ is nilpotent iff all its eigenvalues are 0, but I'm not sure how to prove that $A$ is unipotent iff all of its eigenvalues are 1. For the right to left direction, I use Cayley Hamilton as I did for the right to left direction for the nilpotent statement, but I don't think my strategy for the left to right direction for the nilpotent statement works for the unipotent statement. For the nilpotent statement, I proved the contrapositive and assumed we nonzero eigenvalues. Then, I showed that we have $A^kx = \lambda^kx \neq 0$, so $A^k \neq 0$. But I don't think I can do something similar for unipotent.

Answer 1

Recall:

• The zeroes of the minimal polynomial of $A$ are precisely the eigenvalues of $A$.
• A polynomial $p \in \Bbb{C}[x]$ annihilates $A$ if and only if the minimal polynomial of $A$ divides $p$.

We have \begin{align} A \text{ is unipotent} &\iff (A-I)^n = 0\\ &\iff \text{ the polynomial $(x-1)^n$ annihilates $A$}\\ &\iff \text{ the minimal polynomial of $A$ divides $(x-1)^n$} \\ &\iff \text{ the minimal polynomial of $A$ is of the form $(x-1)^k$ for some $1 \le k \le n$} \\ &\iff \text{ all eigenvalues of $A$ are equal to $1$.} \end{align}