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Consider $n$ real, symmetric, and positive semi-definite matrices as: $A_1,A_2,\cdots,A_n$. These matrices are convertible to each other under appropriate permutation ($A_i(p_i,p_i)=A_j$). Moreover, we have $A_i=Q_iV_iQ^{-1}_i$ because of symmetry. Can we say that $\rho(A_1+A_2+\cdots+A_n )= \rho(V_1+V_2+\cdots+V_n)$ where $\rho(X)$ represents spectral radius of $X$?

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  • $\begingroup$ Hi Mat123, I voted to close your question. For this kind of conjectures, you are expected to do some numerical experiments first to see if the problem statement holds or not. $\endgroup$ – user1551 Aug 18 '15 at 6:27
  • $\begingroup$ Could you please reformulate the confusing statement: These matrices are all identical to each other, if each is permuted by a suitable permutation vector $(A_i(p_i,p_i)=B)$.? I do not find it clear at all. $\endgroup$ – Alex M. Aug 18 '15 at 8:31
  • $\begingroup$ Hi user1551, this question has arisen from a numerical experiment. $\endgroup$ – Mat123 Aug 18 '15 at 16:24
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This is not an answer, just some thoughts

Since all of them are identical to $B$ over permutation, there should be permutation matrices $P_i$ such that $A_i=P_iBP_i^-1$. Also, if $B=QVQ^{-1}$, then $A_i=(P_iQ)V(P_iQ)^{-1}$. Thus if you define $Q_i=P_iQ$, then $A_i=Q_iVQ_i^{-1}$. Therefore, the equality you are seeking can hold only if $Q$ is permutation invariant which in general is not true.

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