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Let $A$ and $B$ be $n \times n$ matrices with $n$ distinct eigenvalues $\lambda_1, \ldots, \lambda_n$ and $\mu_1, \ldots, \mu_n$, respectively.

Is it possible to estimate the difference $|\lambda_i - \mu_i|$ between the eigenvalues of $A$ and $B$ in terms of the matrix norm of $A - B$? If so, which norm should be used, and how can this estimate be determined?

In particular, since all norms are equivalent in finte dimensional space, does $\|A\| \rightarrow 0$ imply $\lambda_i \rightarrow 0$?

Celestina
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    Suppose that I give you two families $(A_t){t\in\Bbb R}$ and $(B_t){t\in\Bbb R}$ of matrices such that $\lim_{t\to\infty}|A_t-B_t|=\infty$ for any norm and that $\max_i|\lambda_i-\mu_i|\leqslant1$. Will that answer your question? – José Carlos Santos Aug 13 '24 at 18:58
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    Eigenvalues are not indicative of the norm for nonsymmetric matrices. $\begin{pmatrix}0 & x \ 0 & 0 \end{pmatrix}$ has a $2$-norm of $|x|$ but its only eigenvalue is $0$. A better version of your question would be to consider the singular values of $A$ and $B$, as these do correlate to the norm – whpowell96 Aug 13 '24 at 19:02
  • @Jose Carlos Santos Thanks for your comment. Actually, I need the other way around. Could you provide an example where ∥A−B∥→0 but the difference in eigenvalues does not go to zero. Anyways, I would be happy to see the example you were mentioning. Thanks in advance – Celestina Aug 13 '24 at 19:04
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    Eigenvalues are continuous functions of matrix coefficients. $|A|\to 0 \implies |a_{ij}|\to 0 \implies \lambda_i\to 0$ – whpowell96 Aug 13 '24 at 19:05
  • @whpowell96 Thanks for your comment. I am looking for the other way round. – Celestina Aug 13 '24 at 19:06
  • @whpowell96 Thanks this is what I am looking for. Could you please elaborate it a bit. Why eigen values are continuous fuicntions of matrix coefficients? Are thy lipschitz continuous? – Celestina Aug 13 '24 at 19:07
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    https://math.stackexchange.com/questions/63196/continuity-of-the-roots-of-a-polynomial-in-terms-of-its-coefficients – whpowell96 Aug 13 '24 at 19:08
  • @whpowell96 Thanks a lot. Ah, it was indeed simple. My bad—I did not think about it – Celestina Aug 13 '24 at 19:10
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    What I had in mind was $n=2$, $A_t=\left[\begin{smallmatrix}2&t\0&0\end{smallmatrix}\right]$, and $B_t=\operatorname{Id}2$. Then, for any norm $|\cdot|$ on the space of all $2\times2$ real matrices, $\lim{t\to\infty}|A_t-B_t|=\infty$. But the eigenvalues of $A_t$ are $2$ and $0$, whereas the only eigenvalue of $B_t$ is $1$. – José Carlos Santos Aug 13 '24 at 19:12

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