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We will prove that if V and W have the same dimension, there exists a linear transformation T: V → W such that T is bijective (injective and surjective), and for all u, v ∈ V, the inner product of T(u) and T(v) is equal to the inner product of u and v. (A) Define a linear transformation T: V → W with a basis {\(v_1, ..., v_n\)} of V and a basis {\(w_1,...,w_n\)} of W, such that T(v_i) = w_i for i = 1, 2,...,n. (B) Show that T is bijective: Injective: Assume there are vectors u, v ∈ V such that T(u) = T(v). We should show that u = v. T(u-v) = T(u) - T(v) = 0, which means u-v is in the kernel of T. Since T is a bijection between V and W, the kernel of T is trivial, so u-v = 0, and therefore, u = v. Surjective: For every vector w ∈ W, we can represent w as a linear combination of the basis vectors {\(w_1,...,w_n\)}, w = c_1 w_1 + ... + c_n w_n. Since T is a bijection between V and W, T(c_1 v_1 + ... + c_n v_n) = c_1 T(v_1) + ... + c_n T(v_n) = c_1 w_1 + ... + c_n w_n = w, so T is surjective. (C) Show that the inner product is preserved: We want to show that for all u, v ∈ V, ⟨T(u), T(v)⟩_W = ⟨u, v⟩_V. Consider two vectors u,v ∈ V such that u = a_1 v_1 + ... + a_n v_n, and v = b_1 v_1 + ... + b_n v_n. Then, ⟨T(u), T(v)⟩_W = ⟨T(a_1 v_1 + ... + a_n v_n), T(b_1 v_1 + ... + b_n v_n)⟩_W = ⟨a_1 w_1 + ... + a_n w_n, b_1 w_1 + ... + b_n w_n⟩_W Since {\(w_1,...,w_n\)} is an orthonormal basis in W, this reduces to: = a_1 b_1 ⟨w_1, w_1⟩_W + ... + a_n b_n ⟨w_n, w_n⟩_W = a_1 b_1 + ... + a_n b_n = ⟨u, v⟩_V Thus, the inner product is preserved, and V and W are isomorphic.

Suppose there is an isomorphism T: V → W. This means that T is linear, injective, and surjective. (A) Show that T maps a basis of V to a linearly independent set in W: Let {\(v_1, ..., v_n\)} be a basis for V. If T(v_i) = 0 for some i, then T is not injective, which is a contradiction. Also, if there is some nontrivial linear combination of T(v_i) that is equal to 0, it would contradict the injective property of T. Thus, the set {\(T(v_1), ..., T(v_n)\)} is linearly independent in W. (B) Show that {\(T(v_1), ..., T(v_n)\)} spans W: Since T is surjective, for any vector w ∈ W, there is a vector v ∈ V such that T(v) = w. But v can be expressed as a linear combination of the basis vectors of V: v = c_1 v_1 + ... + c_n v_n. Thus, w = T(v) = T(c_1 v_1 + ... + c_n v_n) = c_1 T(v_1) + ... + c_n T(v_n). This shows that {\(T(v_1), ..., T(v_n)\)} spans W. (C) Conclude that V and W have the same dimension: Since we've shown that the set {\(T(v_1), ..., T(v_n)\)} is linearly independent and spans W, it forms a basis for W. Thus, the dimension of W is the same as the dimension of V. Now we have proved both directions: (1) If V and W have the same dimension, they are isomorphic, and (2) If V and W are isomorphic, they have the same dimension. This completes the proof.