Let V and W be finitedimensional vector spaces and T V right

Let V and W be finite-dimensional vector spaces and T: V rightarrow W be an isomorphism. If Z is a subspace of V, show that dim(Z) = dim(T(Z)).

Solution

An isomorphism is bijective mapping that preserves sets and relations among elements. Also, in the category of vector spaces, an isomorphism is an invertible linear map.

Since T: V W is an isomorphism, it is bijective so that every element in Z has a distinct image in T(Z), and no two elements in Z have the same image in T(Z) . Also, every element in T(Z) is the image of atleast one ( actually only one unique) element in Z.   Now, let B = { v₁ , v₂ , v₃ , ….,v_n } basis for Z. Then, v₁ , v₂ , v₃ , ….,v_n are linearly independent ( as Z is a vector space) so that no vector in this set B is a linear combination of the other vectors in B. Further, let x = a₁ v₁ + a₂ v₂ + … + a_n v_n be an arbitrary vector in Z. Also let y = T(x). Then y = T(x) = T( a₁ v₁ + a₂ v₂ + … + a_n v_n) = a₁T(v₁)+a₂ T(v₂) +…+a_n T(v_n ) (as T is a linear map being an isomorphism between two vector spaces). Also, the T(v_i) s are linearly independent. Otherwise, if a₁T(v₁)+a₂ T(v₂) +…+a_n T(v_n ) = 0 and all the a_i s are not zero, then T^-1 (a₁T(v₁)+a₂ T(v₂) +…+a_n T(v_n )) = 0 or, a₁ v₁ + a₂ v₂ + … + a_n v_n = 0 which means that v₁ , v₂ , v₃ , ….,v_n are not linearly independent. This is a contradiction. Hence the T(v_i) s are linearly independent. Therefore, { T(v₁ ), T(v₂ ),…, T(v_n )} fors a basis for T(Z). Hence dim(Z) = dim (T(Z)).