# Determine rank of T and whether it is an isomorphism

1. Mar 11, 2017

### hotvette

1. The problem statement, all variables and given/known data
$T((x_0, x_1, x_2)) = (0, x_0, x_1, x_2)$

2. Relevant equations
None

3. 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 $T((x_0, x_1, x_2, \cdots))=(0,x_0, x_1, x_2, \cdots)$ 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?

2. Mar 11, 2017

### Staff: Mentor

$T(x)=(0,x)$ 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.