4 ms·
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
by 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 agoM is hermitian and an (n-1) x (n-1) matrix so that would be reasonable.
- kmill 7y agoThey'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.