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


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?
 
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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.
 

What is the rank of T?

The rank of T is the number of linearly independent rows or columns in the matrix representation of the linear transformation T.

How do you determine the rank of T?

To determine the rank of T, you can row reduce the matrix representation of T and count the number of non-zero rows or columns. Alternatively, you can use the rank-nullity theorem which states that the rank of T is equal to the dimension of the image of T.

What does it mean for T to be an isomorphism?

A linear transformation T is an isomorphism if it is both injective (one-to-one) and surjective (onto). This means that T maps distinct inputs to distinct outputs and that every element in the output space is mapped to by at least one element in the input space.

How can you determine if T is an isomorphism?

To determine if T is an isomorphism, you can check if the nullity (dimension of the kernel) of T is equal to 0. If it is, then T is injective. You can also check if the dimension of the image of T is equal to the dimension of the output space. If it is, then T is surjective. If both conditions are met, then T is an isomorphism.

Can a linear transformation T be an isomorphism if its rank is not equal to the dimension of the input or output space?

No, a linear transformation T can only be an isomorphism if its rank is equal to the dimension of both the input and output space. If the rank is less than the dimension, then T is not surjective. If the rank is greater than the dimension, then T is not injective.

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