# Matrix over $\mathbb{C}$ with non-zero eigenvalues

Let $M$ be a $n \times n$ matrix over the $\mathbb{C}$. Let $E$ be the set of eigenvalues, that is

$$E = \{\lambda \in \mathbb{C}: \exists v \in \mathbb{C}^n\setminus\{0\}, Mv=\lambda v\}.$$

By my previous question:

Does every invertible complex matrix have a non-zero eigenvalue?

i know that if the matrix is invertible it must have non zero eigenvalues, i.e. $E\cap\{0\}^c \neq \emptyset$ , but invertibility is clearly not a necessary condition. So my question is, what is a nice condition weaker then invertibility that is equivalent to $M$ having non-zero eigenvalues.

• No invertibility is not necessary $$\begin{bmatrix}1 & 0 \\ 0& 0\end{bmatrix}$$ Jun 14, 2018 at 11:54
• yes of course, thanks
– john
Jun 14, 2018 at 11:56

This is the case if and only if $M^n\neq 0$.

The reason is that if there is some $m$ such that $M^m = 0$ then also $M^n = 0$ since this is an $n\times n$ matrix. And if $M^m = 0$ for some $m$ then all eigenvalues are $0$, since any eigenvalue $\lambda$ will then also satisfy $\lambda^m = 0$ and thus be $0$.

In the other direction, if all eigenvalues are $0$ then $M$ is conjugate to an upper triangular matrix with $0$s on the diagonal and thus satisfies $M^n = 0$.

• $$M=\begin{bmatrix}1 & 0 \\ 0& 0\end{bmatrix}$$ has $M^2=M$ Jun 14, 2018 at 11:58
• @N8tron Right, so $M^2\neq 0$ and it has a non-zero eigenvalue as I claimed. Jun 14, 2018 at 11:59
• Ah sorry I got a little mixed up there. I thought you were claiming that invertibility was the necessary condition for having a nonzero eigenvectors when $M^n \neq 0$ which is clearly not true :-) Jun 14, 2018 at 12:00
• @N8tron Ahh. My first line answers the question in the title. The question in the body is slightly different (and basically two related questions in one). Jun 14, 2018 at 12:01
• @TobiasKildetoft very nice, thank you!
– john
Jun 14, 2018 at 12:30