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.

  • $\begingroup$ No invertibility is not necessary $$\begin{bmatrix}1 & 0 \\ 0& 0\end{bmatrix}$$ $\endgroup$
    – N8tron
    Jun 14, 2018 at 11:54
  • $\begingroup$ yes of course, thanks $\endgroup$
    – john
    Jun 14, 2018 at 11:56

1 Answer 1


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 $M^n = 0$.

  • $\begingroup$ $$M=\begin{bmatrix}1 & 0 \\ 0& 0\end{bmatrix}$$ has $M^2=M$ $\endgroup$
    – N8tron
    Jun 14, 2018 at 11:58
  • $\begingroup$ @N8tron Right, so $M^2\neq 0$ and it has a non-zero eigenvalue as I claimed. $\endgroup$ Jun 14, 2018 at 11:59
  • $\begingroup$ 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 :-) $\endgroup$
    – N8tron
    Jun 14, 2018 at 12:00
  • 1
    $\begingroup$ @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). $\endgroup$ Jun 14, 2018 at 12:01
  • $\begingroup$ @TobiasKildetoft very nice, thank you! $\endgroup$
    – john
    Jun 14, 2018 at 12:30

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