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He says that eigenvalues of A are the lambda_i(A), so it's pretty obvious that lambda_i(M_j) is the i-th eigen value of M_j. What might not be obvious is the o
by chombier 7y ago
He says that eigenvalues of A are the lambda_i(A), so it's pretty obvious that lambda_i(M_j) is the i-th eigen value of M_j.
What might not be obvious is the ordering of the eigenvalues, which must be the same for the formula to make sense.
- auggierose 7y agoWell, you see, the ordering of the eigenvalues doesn't really matter. What, that's not obvious to you from the statement?
- FabHK 7y agoNo, the ordering of the eigenvalues of M doesn't matter, because we take their product (or rather, the product of their difference from a fixed number), and multiplication commutes. It's maybe not obvious, and as such I can understand auggierose's complaint to an extent, though it was maybe unnecessarily harshly phrased.
- chombier 7y agoYou're right, my bad.