[Math] Characteristic polynomial – using rank

Question

Q:
Let $A$ be an $n\times n$ matrix defined by $A_{ij}=1$ for all $i,j$.
Find the characteristic polynomial of $A$.

There is probably a way to calculate the characteristic polynomial $(\det(A-tI))$ directly but I've spent a while not getting anywhere and it seems cumbersome. Something tells me there is a more intelligent and elegant way. The rank of $A$ is only 2. Is there a way to use this?

Answer 1

1. Notice that $n$ is an eigenvalue with a certain eigenvector.
2. Notice that the rank is 1, so the kernel has dimension $n-1$.

This gives you a basis of eigenvectors ($n-1$ of them have eigenvalue 0).