# Eigenvalues of nilpotent matrices

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!

• The overall claim is that a matrix is nilpotent if and only if all its eigenvalues are zero. – Batman Mar 28 '15 at 12:54
• i don't see any questions. – abel Mar 28 '15 at 12:54
• added the obligatory question, to make things clear – Dahn Mar 28 '15 at 12:55
• and sorry for perhaps a bit nitpicky stream of claims to validate, but I just wanted to make sure my understanding is 100% correct – Dahn Mar 28 '15 at 12:55
• @Batman and the implication $\Leftarrow$ is in these two claims, too? I don't see that. – Dahn Mar 28 '15 at 12:57

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.

• Point 2 is false for $k=0$? I didn't even know there could be $k=0$! What is a $0\times 0$ matrix? Also, what do you mean by "anything implies 1"? – Dahn Mar 30 '15 at 12:16
• @DahnJahn: If you want all linear operators on finite dimensional vector spaces to be representable by square matrices, then you need the (unique) $0\times 0$ matrix for the $0$-dimensional case (and yes, the existence of nonzero vectors is not an axiom of vector spaces). It acts on $\Bbb R^0$, and of course the zero vector of that space is mapped to itself, no problem. This matrix has no eigenvalues (not even complex ones) or eigenvectors, as there are no nonzero vectors to begin with; its characteristic polynomial is $1$. And it is nilpotent (even zero), although it is the identity too. – Marc van Leeuwen Mar 30 '15 at 13:49
• I will have to read a bit more on $\mathbb R^0$! Thanks. – Dahn Mar 31 '15 at 14:16