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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?

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    $\begingroup$ Yes it must have. It is a direct consequence of the Jordan Normal Form. $\endgroup$ – mathreadler Aug 2 '15 at 18:16
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    $\begingroup$ As the answers show, even more is true: every square complex matrix has a nonzero eigenvector. $\endgroup$ – Lee Mosher Aug 2 '15 at 18:29
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    $\begingroup$ Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – Fly by Night Aug 2 '15 at 19:13
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    $\begingroup$ @mathreadler the zero matrix has every eigenvector you can think of ... $\endgroup$ – Gabriel Romon Aug 2 '15 at 19:20
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    $\begingroup$ What would non-zero eigenvector mean except "eigenvector associated with non-zero eigenvalue"? $\endgroup$ – mathreadler Aug 2 '15 at 19:44
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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, which means the matrix has a non-zero eigenvector.

Note that invertibility has nothing to do with the previous reasoning.

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Since $\mathbb{C}$ is algebraically closed, any matrix $A$ can be written (by change of basis) in Jordan normal form. Each Jordan block has exactly one eigenvector associated with it, so there is at least one eigenvector. (Worst case scenario, there is only one block; for example, this is the case with the matrices $\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} i & 1 & 0 \\ 0 & i & 1 \\ 0 & 0 & i \end{bmatrix}$. In this case there is exactly one eigenvector.) We don't need to assume that $A$ is invertible.

For other possibly more elementary proofs, see here and here. Note that having at least one eigenvalue is equivalent to having at least one eigenvector.

Also note, that eigenvectors are by definition nonzero, so "nonzero eigenvector" is weird. Perhaps you mean nonzero eigenvalue.

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$M$ is invertible $\Rightarrow$ $M$ has non zero eigenvalues: proof by contradiction, assume $M$ has only zero eigenvalues, consider the polynomial of order $n$ in $\lambda$

\begin{equation}\det(A-\lambda \mathbb{1}) = \sum_{j=1}^n a_j\lambda^j\end{equation}

which over $ \mathbb{C}$ has a decomposition

\begin{equation}\det(A-\lambda \mathbb{1}) = (\lambda- b_1)\cdot ...\cdot (\lambda - b_n)\end{equation}

now the statement that $M$ has only zero eigenvalues, is equivalent to the above polynomial only having zero as a root, which given the decomposition yields

\begin{equation}\det(A-\lambda \mathbb{1}) = \lambda^n\end{equation}

so in particular for $\lambda = 0$ we get $\det(A) = 0$, which is a contradiction.

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  • $\begingroup$ This answers by original question, however I still don't know if invertibility is a necessary condition $\endgroup$ – john Jun 14 '18 at 11:34
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Let $A\in \Bbb C^{n\times n}$ be invertible $\iff |A|≠0$. Also $|A|=\lambda_1...\lambda_n$, therefore...

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