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

Put content here**Theorem.** A square matrix $A$ is nilpotent if and only if its only eigenvalue is 0.
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**Proof ($\Rightarrow$):** If $A^k = 0$ and $Av = \lambda v$ with $v \neq 0$, then $A^k v = \lambda^k v = 0$, so $\lambda = 0$.
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**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.
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**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 $A^3 = 0$.

# Parents

* Nilpotent matrices