Kunze Linear Algebra · Problem 5.2.5

The key insight is that a nilpotent matrix like $A$ (where $A^2 = 0$) has a very constrained structure—its minimal polynomial divides $x^2$, which severely limits its eigenvalues. Since $\det(cI - A)$ is the characteristic polynomial evaluated at $c$, you need to show that $c$ is a double root, meaning both eigenvalues of $A$ are zero. Use the fact that $A^2 = 0$ to prove $A$ is diagonalizable to a Jordan form with only zero blocks, then compute the determinant directly.

The key insight is that a nilpotent matrix like $A$ (where $A^2 = 0$) has a very constrained structure—its minimal polynomial divides $x^2$, which severely limits its eigenvalues. Since $\det(cI - A)$ is the characteristic polynomial evaluated at $c$, you need to show that $c$ is a double root, meaning both eigenvalues of $A$ are zero. Use the fact that $A^2 = 0$ to prove $A$ is diagonalizable to a Jordan form with only zero blocks, then compute the determinant directly.