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Let $$ A=\left [ \begin{array}{rr} \alpha & \beta\\ -\beta & \alpha\\ \end{array} \right ], $$ where $\alpha, \beta$ are real. Is it possible to find a sequence of real matrices $$ A_n=\left [ \begin{array}{rr} a_n & b_n \\ c_n & d_n \\ \end{array} \right ] $$ with different real eigenvalues such that $\displaystyle\lim_{n \to \infty} A_n=A$? Thanks.

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up vote 6 down vote accepted

The condition "$A_n$ has different real eigenvalues" can be rewritten as $(a_n-d_n)^2+4b_nc_n>0$. If such a sequence exists, we should have $\displaystyle \lim_{n\to +\infty}a_n=\alpha$, $\displaystyle \lim_{n\to +\infty}b_n=\beta$, $\displaystyle \lim_{n\to +\infty}c_n=-\beta$ and $\displaystyle \lim_{n\to +\infty}d_n=\alpha$, hence $(\alpha-\alpha)^2-4\beta^2\geq 0$, which is impossible if $\beta\neq 0$, since $\beta\in\mathbb R$ .

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If $\beta=0$, it's easy. Otherwise, it's impossible. The characteristic polynomial of $A_n$ has to approach that of $A$, so its eigenvalues have to approach those of $A$, and you can't approach a complex number with nonzero imaginary part through the reals.

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