7
$\begingroup$

This question comes from a colleague working in econometrics. $A$ and $B$ are $n\times n$ real symmetric matrices. If we know the eigenvalues of $A$, $B$ and $A+B$, what meaningful information can we obtain about the eigenvalues of $A^2+B^2$?

I have read this related question, but here we have the extra information of the eigenvalues of $A+B$.

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ We may get many inequalities for them, but i am afraid that to describe the complete set of inequalities is rather hopeless. But maybe for applications in econometrics we do need them all? If so, the question should be specified... $\endgroup$
    – Fedor Petrov
    Commented May 25, 2011 at 20:45

1 Answer 1

Reset to default
1
$\begingroup$

This is a partial answer, that could evolve in a few hours.

If $\theta+\nu>0$, then you have $$A^2+B^2\ge(1-\theta)A^2+(1-\nu)B^2+\frac{\theta\nu}{\theta+\nu}(A+B)^2.$$ With Horn's inequalities, you deduce inequalities for the eigenvalues of $A^2+B^2$ in terms of those of $A^2$, $B^2$ and $(A+B)^2$. The latter are the squares of the eigenvalues of $A$, $B$ and $A+B$.

For instance, using Weyl's inequality $$\lambda_k(F+G)\ge\lambda_i(F)+\lambda_j(G),\qquad k+1=i+j,$$ we obtain $$\lambda_m(A^2+B^2)\ge\lambda_i((1-\theta)A^2)+\lambda_j((1-\nu)B^2)+\lambda_k\left(\frac{\theta\nu}{\theta+\nu}(A+B)^2\right),\qquad m+2=i+j+k.$$

Improve this answer
$\endgroup$
2
  • $\begingroup$ @Denis I have voted your answer, but I am still interested in any evolution. $\endgroup$
    – Julián Aguirre
    Commented May 26, 2011 at 20:46
  • $\begingroup$ also note that if $A, B \ge 0$, then a result of Ando and Zhan says that $\lambda(A^r+B^r) \preceq_w \lambda( (A+B)^r)$ for $r\ge 1$ (weak majorization). $\endgroup$
    – Suvrit
    Commented May 29, 2011 at 18:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .