Eigenvalues of nilpotent matrices

Question

I have these two claims for a real $k\times k$ matrix $A$

1 If $A^n=0_{k\times k}$ for some $n\in\mathbb N$ and $\lambda$ is an eigenvalue of $A$, then $\lambda = 0$.

2 If $A^n=0_{k\times k}$ for some $n\in\mathbb N$, then 0 is an eigenvalue of $A$.

I have multiple questions / claims that I want to check:

1. 1) means that all eigenvalues are 0
2. 2) means that at least one of the eigenvalues is 0
3. In that sense, 1) is stronger
4. Both are correct, since 1) yields $\lambda^n=0$ and from the fact that $A$ has to be singular
5. If we allow for complex eigenvalues, then $1) \Rightarrow 2)$

Are these claims correct?

Thank you!

Answer 1

1. Yes, saying that an arbitrary eigenvalues is $0$ means that all eigenvalues, if any, are$~0$.

2. Yes, saying that $0$ is an eigenvalue means that at least one of the eigenvalues is $0$.

3. No this does not mean that 1. implies 2., since there need no be any eigenvalue. It it true that the existence of an eigenvalue together with point 1. implies point 2.

4. Point 1. is correct, from the general fact that any polynomial equation satisfied by a matrix (here $X^n=0$) is also satisfied by any of its eigenvalues. Point 2 is false for $k=0$. If there is any nonzero vector$~v$, then the last nonzero vector among $A^i$ for $i=0,1,\ldots,n$ is an eigenvector for$~0$.

5. As said point 1. is true by itself, so 2. is not needed or helpful, but formally anything implies point 1.