I believe the characteristic polynomial is $p_A(x)=x^n-x^{n-1}$, but I don't know how we get this? I've tried applying the definition $p_A(x)=\text{det}(A-\lambda I)$ but am getting no success with the excessive notation.
2026-03-25 14:24:33.1774448673
How to find characteristic polynomial of a matrix $A$ such that $A=A^n$ for some $n\geq 2$?
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It turns out that there are many possibilities for the characteristic polynomial of $A$. It is true that $x^n - x^{n-1}$ is one such possibility.
Hint: If $A = A^n$, then for every eigenvector $v$ of $A$ we have $Av = A^n v$. What does this tell you about the eigenvalues of $A$?