O I A
... say it is c . The matrix M may have other non- zero entries. Consider the diagonal matrices Di and D j defined as in (5).We have M ij = c.Di .M .D j and therefore M ij belongs to A( ρ ) as claimed, concluding the proof that A( ρ ) = A . The two claims above being proved, the isomorphism stated in th ...
... say it is c . The matrix M may have other non- zero entries. Consider the diagonal matrices Di and D j defined as in (5).We have M ij = c.Di .M .D j and therefore M ij belongs to A( ρ ) as claimed, concluding the proof that A( ρ ) = A . The two claims above being proved, the isomorphism stated in th ...
Linear Equations - MDC Faculty Web Pages
... An equation consists of two algebraic expressions joined by an equal sign. 3x – 1 = 2 3x = 3 x=1 ...
... An equation consists of two algebraic expressions joined by an equal sign. 3x – 1 = 2 3x = 3 x=1 ...
SOME PROPERTIES OF N-SUPERCYCLIC OPERATORS 1
... normal operator on an infinite-dimensional Hilbert space may be N -supercyclic. Observe that if an operator T is hypercyclic or supercyclic, then it is clearly cyclic. However, an N -supercyclic operator need not be cyclic—trivial examples are furnished by the observation that if dim(X ) = N , then ...
... normal operator on an infinite-dimensional Hilbert space may be N -supercyclic. Observe that if an operator T is hypercyclic or supercyclic, then it is clearly cyclic. However, an N -supercyclic operator need not be cyclic—trivial examples are furnished by the observation that if dim(X ) = N , then ...
Mar 2006 Selected Problems, Chapter 3 Math 230(Mackey) Revised
... equalling the zero vector. This says that S is a linearly dependent set. But this contradicts what we were given! So our assumption that T is a linearly dependent set is false. Hence T is a linearly independent set. Since T was an arbitrary subset of S, we conclude that every subset of a linearly in ...
... equalling the zero vector. This says that S is a linearly dependent set. But this contradicts what we were given! So our assumption that T is a linearly dependent set is false. Hence T is a linearly independent set. Since T was an arbitrary subset of S, we conclude that every subset of a linearly in ...
