3 ms·
It’s equivalent to the usual definition, but spares you a few steps when proving the existence of eigenvalues. I.e. you only need to prove that T - hI maps some
by tzahola 9y ago
It’s equivalent to the usual definition, but spares you a few steps when proving the existence of eigenvalues. I.e. you only need to prove that T - hI maps some nonzero vector to zero.