# Is there an open ball around the $0$ matrix such that the eigenvalues contains a cube?

Let $A \in \mathcal M(n \times n; \mathbb C$. We know the eigenvalues of a matrix are continuous functions of the entries of a matrix. Let us define a map $F$ as follows \begin{align*} F: A \mapsto (\lambda_1(A), \dots, \lambda_n(A)) \mapsto (|\lambda_1(A)|, \dots, |\lambda_n(A)|), \end{align*} where $\lambda_j(\cdot)$ denotes the eigenvalue function and $|\cdot|$ denote the modulus. $F$ is continuous function into $\mathbb R_+^n$. Let us fix a norm $\|\cdot\|$ on $\mathcal M_n(\mathbb C)$. I am wondering whether or not for every open ball $B_0(\varepsilon)$ around the zero matrix, the image under $F$, $F[ B_0(\varepsilon) ]$ contains a cube in $\mathbb R_+^n$, i.e., $[0, \varepsilon')^n$ for some probably smaller $\epsilon'$?

• The ball contains suitable diagonal matrices ... – Hagen von Eitzen Jul 7 '18 at 21:55
• Of course. How silly I didn't consider that. Thanks. – user1101010 Jul 7 '18 at 21:57
• @HagenvonEitzen: Is it true for a general element $A$? If it is diagonalizable, I think we can use the hint in your comment. What if it is not diagonalizable? – user1101010 Jul 7 '18 at 22:00