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Let $(A_n)_{n\in\mathbb{N}}$ be a sequence of complex hermitian $m\times m$ matrices that converges to another complex hermitian $m\times m$ matrix $A$ in the Frobenius norm, i.e. $$\lim_{n\to\infty}\|A-A_n\|_{F}=0.$$ For $n\in\mathbb{N}$ let $\lambda^{(1)}_n\ge\cdots\ge\lambda^{(m)}_n$ be the ordered eigenvalues (they are real because of the spectral theorem) of $A_n$ and let $\lambda^{(1)}\ge\cdots\ge\lambda^{(m)}$ be the ordered eigenvalues of $A$.

Question: Do the eigenvalues converge in the sense that for each $j\in\{1,\dots,m\}$ $$\lim_{n\to\infty} \lambda^{(j)}_n=\lambda^{(j)}\quad\text{?}$$

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    And what do you think? Hint: equivalence of norms in finite-dimensioned spaces + continuous dependence of polynomial roots on the coefficients of this polynomial. – 2017-02-22
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    @TZakrevskiy The characteristic polynomials converge pointwise and hence the coefficients in any polynomial basis for degree $m$, but what do the roots do? – 2017-02-22
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    @TZakrevskiy They converge. I didn't know about this continuous dependence before. Thanks very much! – 2017-02-22

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