Examples

"algebraically determined"

Used in print

First , T may not have a single characteristic value ; this is really a deficiency in the scalar_field , namely , that it is not algebraically closed .

Let V be a finite dimensional vector space over an algebraically closed field F , e.g. , the field of complex_numbers .

From these results , one sees that the study of linear_operators on vector spaces over an algebraically closed field is essentially reduced to the study of nilpotent operators .

For vector spaces over non algebraically closed fields , we still need to find some substitute for characteristic values and vectors .

Related terms

algebraic algebraic