Question

Suppose is a subspace of such that / is finite-dimensional. Prove that is isomorphic to × (/)

Answer 1

The subspace V of a finite-dimensional vector space W is isomorphic to
`\(F^n\),` where F is the field of scalars and n is the dimension of V.

Step-by-step explanation:

Let V be a subspace of the finite-dimensional vector space W such that
`\(\dim(V) = n\)`. We aim to prove that V is isomorphic to
`\(F^n\),`where F is the field of scalars.

To establish the isomorphism, consider the map
`\(\phi: F^n \rightarrow V\)`defined by associating each vector in
`\(F^n\)`with its unique representation in V. This map is surjective as each element in V has a preimage in
`\(F^n\).` Additionally,
`\(\phi\)` is injective, as distinct vectors in
`\(F^n\)` map to distinct vectors in V.

By the Rank-Nullity Theorem, since
`\(\dim(\text{ker}(\phi)) = \dim(F^n) - \dim(\text{im}(\phi))\), and \(\dim(\text{ker}(\phi)) = 0\) (as \(\phi\) is injective),`it follows that
`\(\dim(\text{im}(\phi)) = n\)`. Hence,
`\(\phi\)` is an isomorphism between
`\(F^n\)`and V, demonstrating that the subspace V is isomorphic to the direct product
`\(F^n\)`.

In conclusion, the isomorphism between V and
`\(F^n\)` provides a one-to-one correspondence between the vectors in V and
`\(F^n\)`, emphasizing the structural similarity between the subspace and the direct product space.