3 ms·
Yes except eigenvalues need not be 1. e^(kx) is an eigenvector of d/dx for all k but is only a fixed point when k=1
by bobbylarrybobby 4y ago
Yes except eigenvalues need not be 1. e^(kx) is an eigenvector of d/dx for all k but is only a fixed point when k=1
- ukj 4y agoRight, in topology a “point” can represent any mathematical object e.g 1 of “e^(kx)”