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Is there an expression for the largest eigenvalue of the sum of two Hermitian matrices in terms of the spectrum of the same matrices?

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    $\begingroup$ Special case of mathoverflow.net/questions/4224/eigenvalues-of-matrix-sums $\endgroup$
    – Terry Tao
    Sep 1 '14 at 16:38
  • $\begingroup$ @ terry tao. Thanks for pointer. I read your interesting blog post on this matter. However, I was hoping for a closed form in the 2x2 and/or 4x4 cases. Does such a thing exist. Also, I'm only interested in the largest eigenvalue. $\endgroup$
    – Benjamin
    Sep 1 '14 at 16:58
  • $\begingroup$ Actualy, scratch some of that as the 2x2 case is simple $\endgroup$
    – Benjamin
    Sep 1 '14 at 17:09
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    $\begingroup$ This question appears to be off-topic because it is about under-graduate level matrix theory $\endgroup$ Oct 8 '14 at 14:20
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If $A$ has eigenvalues $\lambda_1 \geq \dots \geq \lambda_n$ and $B$ has eigenvalues $\mu_1 \geq \dots \geq \mu_n$, then the largest eigenvalue of $A+B$ can take any value between $\lambda_1+\mu_1$ (the maximal value) and $\max_{1 \leq i \leq n} \lambda_i+\mu_{n+1-i}$ (the minimal value). That these inequalities are necessary comes from the Weyl inequalities (or from the Courant-Fisher-Weyl min-max principle); the extremals are attained when $A,B$ are diagonal with entries in either increasing or decreasing order; and the intermediate cases are then attainable thanks to the intermediate value theorem. (One could also use the Atiyah-Guillemin-Sternberg convexity theorem here, or my work with Allen Knutson, but this is overkill, to put it mildly.)

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  • $\begingroup$ what is the intermediate value theorem in this specific case? $\endgroup$
    – Mr.
    Sep 2 '14 at 7:43
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    $\begingroup$ If a continuous function $f: X \to {\bf R}$ on a connected domain $X$ attains the values $a$ and $b$, then it also attains all intermediate values between $a$ and $b$. (In the case when $X$ is path-connected, which is the case here, this version of the intermediate value theorem can be derived from the classical one, but it can be proven directly without much difficulty from basic point-set topology.) $\endgroup$
    – Terry Tao
    Sep 2 '14 at 16:07
  • $\begingroup$ @TerryTao Can you may be kindly help at this related question? mathoverflow.net/questions/187850/… $\endgroup$
    – Anirbit
    Nov 23 '14 at 0:19

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