Does every invertible complex matrix have a non-zero eigenvalue? Let $M$ be an $n \times n$ matrix over the field of complex numbers. Additionally assume that $M$ is invertible. Now let $E$ be the set of eigenvalues that is

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

Now is it true, that $E\cap\{0\}^c \neq  \emptyset$, i.e. does $M$ always have a non zero eigenvalue.

My thought are that by the fundamental theorem of algebra, I know that every complex polynomial, i.e. every polynomial with complex coefficients has at least one solution, and thus I can conclude that $E\neq \emptyset$. Now the general statement is clearly false if we don't assume that $M$ is invertible, since the zero matrix has only zero as an eigenvalue. So is the condition of invertibility sufficient?
Since every polynomial has a root over $\mathbb{C}$, the characteristic polynomial of any complex matrix must have a root, say $\lambda$. Then $\lambda$ is an eigenvalue of the matrix at hand. Since the matrix is assumed to be invertible, we have $\lambda \neq 0$. 

Regarding the last statement, if $M$ has $0$ as eigenvalue, there is some non-zero eigenvector $x$: $Mx=0$, and $M$ is not invertible.
Let $A\in \Bbb C^{n\times n}$ be invertible $\iff |A|≠0$. Also $|A|=\lambda_1...\lambda_n$, therefore...