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A matrix is nilpotent if and only if its only eigenvalue is 0.

Created over 8 years ago, updated 10 days ago

Theorem. A square matrix $A$ is nilpotent if and only if its only eigenvalue is 0.

Proof ($\Rightarrow$): If $A^k = 0$ and $Av = \lambda v$ with $v \neq 0$, then $A^k v = \lambda^k v = 0$, so $\lambda = 0$.

Proof ($\Leftarrow$): If 0 is the only eigenvalue, the characteristic polynomial is $\det(\lambda I - A) = \lambda^n$. By the Cayley-Hamilton theorem, $A^n = 0$, so $A$ is nilpotent.

Example. $A = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$ has only eigenvalue 0 (upper triangular with zeros on diagonal), so $A^3 = 0$.