3 ms·

In that formula the order of the eigenvalues doesn't matter...

• auggierose 7y ago
  The order of the eigenvalues does not really matter in that formula, no. Thank you for decoding that for me. Still, how many lambda_i(M_j) are there? From the formula I am led to believe there are n-1 such eigenvalues of M_j.
  • banachtarski 7y ago
    M is hermitian and an (n-1) x (n-1) matrix so that would be reasonable.
  • kmill 7y ago
    They're expecting you to know the spectral theorem for Hermitian matrices. Since M_j is obtained by striking out the jth row and jth column, it too is Hermitian, so it has has n-1 real eigenvalues (possibly with multiplicity). The notation $\lambda_i(M)$ ought to refer to the $i$th eigenvalue of M in sorted order to be unambiguous, but this does not matter for the proof.