Assume we have a density operator (Hermitian, PSD, with trace 1, where PSD means positive semi-definite) called A for a particle. $v_i$ shows the i-th eigenvector of A and $\lambda_i$ shows the i-th eigenvalue. I assume that the elements of $|v_{i}|^2$ form a probability distribution of the particle being in the state $i$, and $\lambda_i$ shows the probability of happening the i-th state. Now if we omit the j-th row and column of A (called $A_{-j}$, is there any intuition behind the obtained matrix and its eigenvalues?


When you remove a column and corresponding row from a matrix, the eigenvalues and eigenvectors change according to this: https://mathoverflow.net/questions/43514/how-do-eigenvectors-and-eigenvalues-change-when-we-remove-a-row-column-pair-of-a

The only additional thing I will say here is what the consequences would be in quantum mechanics (because the link I gave you is only for general Maths). The consequence would be that what you have may no longer be a density matrix, unless you adjust the rest of the row and columns to compensate for the elements that were taken out (because the sum of the diagonals has to equal 1, so unless the diagonal element deleted was a 0, then you will be affecting the sum of the diagonals when removing that element). If the matrix is no longer a density matrix, then it is no longer a quantum state, so you have to be careful.

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  • $\begingroup$ It is an unnormalized density matrix, though (ie positive semidefinite). And can be seen as the state after a measurement which discrimnates if you have a state |i> or are in the subspace orthogonal to it. $\endgroup$ – Norbert Schuch Aug 23 '19 at 22:32

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