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An isomorphism T between two vector spaces V and W is a linear transformation with the following properties: 1. T is one-to-one (injective): For any distinct vectors u and v in V, their images in W are also distinct, i.e., if u ≠ v, then T(u) ≠ T(v). 2. T is onto (surjective): For any vector w in W, there exists a vector v in V such that T(v) = w.

As given in the problem statement, dim V = n. A basis for V consists of n linearly independent vectors \( \{v_1, v_2, \dots, v_n \} \). We need to create a basis for W with the same number of vectors (dim W = n). Let's consider the image of each vector in the basis of V under T: \( w_i = T(v_i) \) for i = 1, 2, ..., n We will now show that these n vectors \( \{ w_1, w_2, \dots, w_n \} \) form a basis for W.

To demonstrate that the set \( \{ w_1, w_2, \dots, w_n \} \) is linearly independent in W, we need to show that if there is a linear combination of the vectors equals the zero vector in W, then all coefficients must be zero. Consider a linear combination of these vectors: \( c_1w_1 + c_2w_2 + \dots + c_nw_n = 0_W \) where \(c_i\) are scalar coefficients, and \(0_W\) is the zero vector in W. Since T is a linear transformation, we can also write the linear combination as follows: \( T(c_1v_1 + c_2v_2 + \dots + c_nv_n) = 0_W \) Since T is injective (one-to-one), this implies: \( c_1v_1 + c_2v_2 + \dots + c_nv_n = 0_V \) As the vectors \( \{v_1, v_2, \dots, v_n \} \) form a basis for V, they are linearly independent. Therefore, the coefficients must be zero: \( c_1 = c_2 = \dots = c_n = 0 \) This proves that the vectors in \( \{ w_1, w_2, \dots, w_n \} \) are linearly independent.

To show that the set \( \{ w_1, w_2, \dots, w_n \} \) spans W, we need to prove that for any vector w in W, there exists a linear combination of the vectors in the set, making it equal to w. Since T is surjective (onto), for any w in W, there exists a v in V such that: \( w = T(v) \) Now, v can be expressed as a linear combination of the vectors in the basis for V: \( v = a_1v_1 + a_2v_2 + \dots + a_nv_n \) Applying T, we get: \( w = T(a_1v_1 + a_2v_2 + \dots + a_nv_n) \) By linearity of T: \( w = a_1T(v_1) + a_2T(v_2) + \dots + a_nT(v_n) \) \( w = a_1w_1 + a_2w_2 + \dots + a_nw_n\) Thus, we have expressed w as a linear combination of the vectors in the set \( \{ w_1, w_2, \dots, w_n \} \), which means the set spans W.

Since we have proven that \( \{ w_1, w_2, \dots, w_n \} \) is a basis for W, this implies that dim W = n. Therefore, we have shown that an isomorphism between two vector spaces preserves their dimensions, as required: \( \operatorname{dim} V = \operatorname{dim} W = n \)