Determine rank of T and whether it is an isomorphism

In summary, the conversation discusses the definition of an isomorphism and the conditions for a linear map to be an isomorphism. The example given shows that a linear and invertible map can still fail to be an isomorphism if the dimensions of the domain and codomain do not match. It is also noted that a linear map on an infinite sequence can be an isomorphism, even though the dimension test does not apply in this case. However, it is more accurate to say that the map is an isomorphic embedding or a monomorphism.
  • #1
hotvette
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Homework Statement


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

Homework Equations


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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?
 
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  • #2
##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.
 
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