For each m ∈ { 1 , 2 , . . . , n } , is there a transformation ϕ m that I can apply to a matrix M ∈ R n × n such that the m t h largest eigenvalue λ m of M is the smallest eigenvalue λ n ′ of the matrix M ′ = ϕ m ( M )?

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For each   m  ∈  {  1  ,  2  ,  .  .  .  ,  n  } , is there a transformation       ϕ    m   that I can apply to a matrix   M  ∈        R          n      ×      n       such that the       m          t      h       largest eigenvalue       λ    m   of   M is the smallest eigenvalue       λ    n    ′   of the matrix       M    ′    =      ϕ    m    (  M  )?

Brendon Fernandez · Accepted Answer

In absolute value or not ? Because for       M    ′    =  −  M you get to invert the ordering on eigenvalues... If kernel is empty, you can take       M    ′    =      M          −      1       it transforms   λ in       1    λ  . But in both these examples we do not have       λ    1    =      λ    n    ′   , just some relation between them. Do you want equality for just this one eigenvalue ?Note that if you know       λ    1   then       M    ′    =  2      λ    1    I  −  M works. Maybe if you are able to have an upper bound m for the eigenvalues, then       M    ′    =  2  m  I  −  M would also suits your needs (for an algorithm for instance).

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