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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.

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    $\begingroup$ 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. $\endgroup$ – user8268 Nov 3 '18 at 7:30
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    $\begingroup$ @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 :-) $\endgroup$ – Jyrki Lahtonen Nov 3 '18 at 7:42
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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.

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  • $\begingroup$ 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. $\endgroup$ – user1101010 Nov 3 '18 at 17:54

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