B u1 u2 un is called an orthonormal set whenever ui 1 for

B = {u_1, u_2, ..., u_n} is called an orthonormal set whenever ||u_i|| = 1 for each i, and u_i u_j for all i notequalto j. In other words, (u_i|u_j) = {1 when i = j, 0 when i notequalto j. Every orthonormal set is linearly independent. Every orthonormal set of n vectors from an n-dimensional space v is an orthonormal basis for V.

Solution

a).Let us suppose that {v₁, . . ., v_k} is an orthonormal set, and that there are constants c₁, . . ., c_k such that c₁v₁+ c₂v₂+ · · · + c_kv_k = 0. Then, for any i between 1 and k we have v_i.( c₁v₁+ c₂v₂+ · · · + c_kv_k) = 0 Now,          {v₁, . . ., v_k} is an orthonormal set so that v_i.v_j = 0 for i j . Then c_i ||v_i ||² = 0. Further, since v_i 0, (otherwise the set could not be orthonormal), hence c_i = 0. Thus the only linear combinations of vectors in the set {v₁, . . ., v_k} which equal the 0 vector are those in which all the coefficients are zero. This implies that the set {v₁, . . ., v_k} is linearly independent.

(b), Let us suppose that {v₁, . . ., v_n} is an orthonormal set of vectors in an n-dimensional vector space V. Then, by part (a) above, the vectors v₁, . . ., v_n are linearly independent. Further, we know that any set of n linearly independent vectors in an n-dimensional space forms a basis of V. Hence the set {v₁, . . ., v_n} is an orthonormal basis for V.