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This is a well-known result in linear algebra which says that for any inner product space, you can always construct an orthonormal basis. This property is often a result of a process known as Gram-Schmidt orthonormalization. Here's how you can prove it: Given a basis {v1, ..., vn} for V, we want to construct an orthonormal basis {u1, ..., un} for V. We can do this through a process called Gram-Schmidt orthonormalization: Let u1 = v1/||v1||, where ||v|| denotes the norm of a vector v with respect to the given inner product. For each k > 1, define a new vector w_k = v_k - Σ_{j=1}^{k-1} ⟨v_k, u_j⟩ u_j. This process subtracts off the components of v_k in the direction of the previously constructed u's. Let u_k = w_k/||w_k||. The set {u1, ..., un} so constructed is an orthonormal basis for V. Indeed, they are orthogonal by construction, and they are normal because we divide by the norm at each step. They are a basis because they span the same subspace as the v's: the space spanned by the fi
This is a well-known result in linear algebra which says that for any inner product space, you can always construct an orthonormal basis. This property is often a result of a process known as Gram-Schmidt orthonormalization. Here's how you can prove it: Given a basis {v1, ..., vn} for V, we want to construct an orthonormal basis {u1, ..., un} for V. We can do this through a process called Gram-Schmidt orthonormalization: Let u1 = v1/||v1||, where ||v|| denotes the norm of a vector v with respect to the given inner product. For each k > 1, define a new vector w_k = v_k - Σ_{j=1}^{k-1} ⟨v_k, u_j
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