# Does a norm ball around a real matrix contain all possible pairs of complex spectrum?

We consider the normed vector space $$(M_n(\mathbb R), \|\cdot\|_F)$$, i.e., real matrices with Frobenius norm. Let $$A \in M_n(\mathbb R)$$ be a diagonalizable matrix and have all eigenvalues to be real. Let $$B_A(\varepsilon)$$ denote the open norm ball with radius $$\varepsilon > 0$$, i.e., \begin{align*} B_A(\varepsilon) =\{ E \in M_n(\mathbb R): \|A-E\|_F < \varepsilon\}. \end{align*} We know the complex eigenvalues of a real matrix must be conjugate pairs. My question is: for any combination of real or complex conjugate pairs of eigenvalues, is there always a matrix $$E \in B_A(\varepsilon)$$ has the spectrum with the same number of real and complex conjugate pairs. To be more clear, let $$n = 2k$$ be even, I would like to know whether there are always matrices in the norm ball such that have eigenvalues with $$1$$ complex conjugate pair, $$2$$ pairs, and so on until $$k$$ pairs of conjugate eigenvalues.

• If the eigenvalues of $A$ are real and distinct then so are the eigenvalues of any nearby matrix (if you deform a bit the coefficients of a real polynomial with all roots real and distinct then the deformed polynomial has the same property, say by the implicit function theorem). So the answer is no. – user8268 Nov 3 '18 at 7:30
• @user8268 Why don't you flesh that out to an answer!! It is so much simpler than my argument. Not needing an orthogonal basis of eigenvectors and all that :-) – Jyrki Lahtonen Nov 3 '18 at 7:42

If the eigenvalues are distinct, then nearby matrices must also have $$n$$ distinct real eigenvalues as observed by user8268 in the comments.
On the other hand, if the eigenvalues of $$A$$ are not distinct, then there can be a complex conjugate pair of eigenvalues near $$A$$. For example, $$\lVert I_{2\times 2}-R_\theta\rVert = \left\lVert\begin{pmatrix} 1-\cos\theta & \sin\theta\\ -\sin\theta & 1-\cos\theta \end{pmatrix}\right\rVert$$ which is $$\approx C\lvert\theta\rvert$$ for $$\lvert\theta\rvert$$ small. So you can afford to get up to $$m$$ complex conjugate pairs, where $$m=\sum_\lambda\left\lfloor\frac12\operatorname{mult}_\lambda(A)\right\rfloor$$ and not any further.
• Thanks for answering the question. Could you provide an argument for the second part? In particular, does the result depend on the diagonalizability of $A$? Does it still hold if $A$ is not diagolizable? Thanks. – user1101010 Nov 3 '18 at 17:54