Why are eigenvalues of nilpotent matrices equal to zero? If $A$ is a $ \displaystyle  10 \times 10 $ matrix such that $A^{3} = 0$ but $A^{2}  \neq 0$ (so A is nilpotent) then I know that $A$ is not invertible, but why does at least one eigenvalue of $A$ have to be equal to zero? How would one show that all eigenvalues of $A$ are equal to zero?
 A: Alternatively, do the contrapositive. If $A$ has a non-zero eigenvalue, $\lambda,$ then $A^{k} \neq 0 $ for all $k.$ 
Proof:
there exists $v \neq 0,$ such that 
$$
Av = \lambda v,
$$
so
$$
A^{k} v = \lambda^{k} v \neq 0.
$$
Done.
A: If $v$ is a non-zero eigen vector corresponding to an eigenvalues $\lambda$ we have, by definition, $Av=\lambda v$. Then $A^2v= A( Av)= A(\lambda v)= \lambda^2v$. It easily follows that $\lambda^n$ is an eigenvalue for $A^n$ but the latter is the zero matrix, for which all eigenvalues are zero,   hence $\lambda=0$.
A: Suppose $n$ is the smallest integer for which $A^n=0$. Since $A$ is non-zero $\implies$ $n \gt 1$
let the characteristic polynomial for $A$ be
$$
A^m+c_1A^{m-1}+\cdots + c_m = 0
$$
here $m \le n$. multiplying the equation by $A^{n-1}$ gives
$$
c_mA^{n-1} =0
$$
hence $c_m=0$. this procedure can be repeated to show that all coefficients except the first are zero, hence the equation is simply:
$$
A^m=0
$$
and we must have $m=n$
since the eigenvalues are roots of the equation:
$$
x^m = 0
$$
it follows that they must all be zero
A: Suppose $\lambda$ is an eigenvalue of the nilpotent matrix $A,$ and $u$ its associated eigenvector. Then $$Au = \lambda u, u \neq 0$$ multiplying by $A$ on the right shows $$A^2u = \lambda Au = \lambda^2 u$$ and by induction $$A^n u = \lambda^n u$$ 
If $A$ is nilpotent, then $A^k = 0$ for some $k>0.$  that implies $\lambda^k = 0\to \lambda = 0.$
