Question

For a positive integer, define by = × ⋯ × ⏟ times. Prove that and ℒ( , ) are isomorphic vector spaces

Answer 1

The vector spaces
`\( \mathbb{N} \) and \( \mathcal{L}(\mathbb{N}, \mathbb{R}) \)`are isomorphic.This isomorphism is established through a bijective linear map
`\( T: \mathbb{N} \rightarrow \mathcal{L}(\mathbb{N}, \mathbb{R}) \)` defined by
`\( T(n) \)` mapping to the linear transformation
`\( L_n: \mathbb{N} \rightarrow \mathbb{R} \)` with
`\( L_n(f) = n \cdot f \).`

Step-by-step explanation:

To show that
`\( \mathbb{N} \) and \( \mathcal{L}(\mathbb{N}, \mathbb{R}) \)` are isomorphic vector spaces, we need to establish a bijective linear map between them. Let
`\( T: \mathbb{N} \rightarrow \mathcal{L}(\mathbb{N}, \mathbb{R}) \)` be defined by
`\( T(n) \)` mapping to the linear transformation
`\( L_n: \mathbb{N} \rightarrow \mathbb{R} \)` defined by
`\( L_n(f) = n \cdot f \)` for all ( f ) in
`\( \mathbb{N} \).`

The map ( T ) is linear because
`\( T(c_1n_1 + c_2n_2) = c_1T(n_1) + c_2T(n_2) \)` for any scalars
`\( c_1, c_2 \)` and natural numbers
`\( n_1, n_2 \)`. Moreover, ( T ) is injective, as distinct natural numbers map to distinct linear transformations. To show surjectivity, consider any linear transformation ( L ) in
`\( \mathcal{L}(\mathbb{N}, \mathbb{R}) \)`. Define
`\( n = L(1) \).` Then,
`\( T(n) = L \)`, confirming surjectivity.

Thus, we have a bijective linear map between
`\( \mathbb{N} \)` and
`\( \mathcal{L}(\mathbb{N}, \mathbb{R}) \)`, establishing the isomorphism between these vector spaces. This means they share the same structure and can be considered essentially the same from a linear algebra perspective.