The $k^{\rm th}$ largest eigenvalue (arranged in decreasing order) of the sum of two $N \times N$ Hermitian (real symmetric) matrices $\bf{A}$ and $\bf{B}$ can be stated using the Weyl inequalities as

$L_k \leq \lambda_k({\bf A} + {\bf B}) \leq U_k$

with the lower and upper bounds given by

$L_k = {\rm max}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=k+N$

$U_k = {\rm min}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=k+1$.

The bounds on the largest eigenvalue $\lambda_1$ of the matrix sum ${\bf A}+{\bf B}$ are therefore

$L_1 = {\rm max}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=1+N$

$U_1 = {\rm min}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=2$.

Finding the upper bound is very simple; $U_1 = \lambda_1({\bf A})+\lambda_1({\bf B})$ since there is only one solution to the index equation $i+j=2$. However, finding the lower bound is much more difficult; the maximum of

$\left[\lambda_1({\bf A})+\lambda_N({\bf B}), \lambda_2({\bf A})+\lambda_{N-1}({\bf B}), \ldots, \lambda_N({\bf A})+\lambda_1({\bf B}) \right]$

must be found.

Are there any simplifications or theorems that can be applied to obtain a simpler expression for the lower bound (without restricting $\bf B$ to be a slight variation or permutation to $\bf A$) ?


1 Answer 1


Weyl's inequalities are not the full story. The characterization of the possible spectra of $A+B$, given the spectra of $A$ and $B$ is the object of A. Horn's conjecture. This is now a theorem, after hard works by Fulton, Klyachko, Knutson, Terry Tao and others. The conjecture consists in linear inequalities (the simplest of these being Weyl's) which are found recursively on the dimension $n$. After Weyl's inequalities, there are Ky Fan or Wielandt's inequalities, etc ...

  • $\begingroup$ I found this to be a nice introduction to the topic. ams.org/notices/200102/fea-knutson.pdf $\endgroup$ Commented May 25, 2012 at 3:30
  • $\begingroup$ The name is spelled in English as Klyachko. $\endgroup$
    – Misha
    Commented May 27, 2012 at 3:44
  • $\begingroup$ @Misha. I apologize, and edit. $\endgroup$ Commented May 27, 2012 at 14:43

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