Determine rank of T and whether it is an isomorphism

  • Thread starter Thread starter hotvette
  • Start date Start date
  • Tags Tags
    Isomorphism rank
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
Homework Helper
Messages
1,001
Reaction score
11

Homework Statement


[itex]T((x_0, x_1, x_2)) = (0, x_0, x_1, x_2)[/itex]

Homework Equations


None

The Attempt at a Solution


I'm getting hung up on definitions. My book says that T is an is isomorphism if T is linear and invertible. But it goes on to say that for T of finite dimension, T can only be an isomorphism if dim(T(M)) = dim(M). The T as stated is linear and invertible, but dim(T(M)) = 4 and dim(M) = 3 which indicates it isn't an isomprphism. Rank = dim(M) = 4.

If I were to define T as [itex]T((x_0, x_1, x_2, \cdots))=(0,x_0, x_1, x_2, \cdots)[/itex] the dim test doesn't apply because T deals with an infinite sequence, and we can say T is an isomorphism.

It doesn't make sense that in the finite case, T isn't an isomorphism but it is in the infinite case. What am I doing wrong?
 
Physics news on Phys.org
##T(x)=(0,x)## is an isomorphic embedding, a monomorphism, i.e. isomorphic to the image of ##T##. It's just a bit sloppy to say ##T## is an isomorphism, since it is only on a subspace of the codomain.