# Characteristic polynomial - using rank?

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?

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What makes you say the rank is $2$? –  EuYu Nov 1 '12 at 21:52

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).
Actually the rank is $1$, not $2$. Yes, there is a much better way: find the nonzero eigenvalue. Hint: the eigenvector is $(1,\ldots,1)$.