To prove that if two finite-dimensional vector spaces V and W over a field F are isomorphic, then they have the same dimension, we can use the following steps: Step 1: Assume that V and W are isomorphic vector spaces over the field F. Step 2: Since V and W are isomorphic, there exists an isomorphism T: V → W, which is a bijective linear transformation. Step 3: Let {v1, v2, ..., vn} be a basis for V, where n is the dimension of V. Step 4: Since T is an isomorphism, the set {T(v1), T(v2), ..., T(vn)} must be a basis for W. Proof: (a) Linearly independent in W: Suppose there exist scalars α1, α2, ..., αn in F, not all zero, such that α1T(v1) + α2T(v2) + ... + αnT(vn) = 0 (the zero vector in W). Since T is a linear transformation, we have: T(α1v1 + α2v2 + ... + αnvn) = 0 However, since {v1, v2, ..., vn} is a basis for V, the linear combination α1v1 + α2v2 + ... + αnvn cannot be equal to the zero vector in V unless all scalars α1, α2, ..., αn are zero. This contradicts the assumption that not all scalars are zero. Therefore, {T(v1), T(v2), ..., T(vn)} is a linearly independent set in W. (b) Spans W: Let w be any vector in W. Since T is a surjective isomorphism, there exists a vector v in V such that T(v) = w. As {v1, v2, ..., vn} is a basis for V, we can express v as a linear combination of the basis vectors: v = β1v1 + β2v2 + ... + βnvn, for some scalars β1, β2, ..., βn in F. Applying T to both sides, we get: T(v) = T(β1v1 + β2v2 + ... + βnvn) = β1T(v1) + β2T(v2) + ... + βnT(vn) Substituting w = T(v), we have: w = β1T(v1) + β2T(v2) + ... + βnT(vn) Therefore, {T(v1), T(v2), ..., T(vn)} spans W. Step 5: Since {T(v1), T(v2), ..., T(vn)} is a basis for W, the dimension of W is also n. Therefore, we have proved that if V and W are isomorphic finite-dimensional vector spaces over a field F, then they must have the same dimension.