By Edwards, Charles Henry

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An be an orthonormal basis for n. If x = s1 a1 + · · · + sn an and y = t1 a1 + · · · + tn an, show that x · y = s1t1 + · · · + sn tn. That is, in computing x · y, one may replace the coordinates of x and y by their components relative to any orthonormal basis for n. 7 Orthogonalize the basis (1, 0, 0, 1), (−1, 0, 2, 1), (0, 1, 2, 0), (0, 0, −1, 1) in 4. 8 Orthogonalize the basis in n. 9 Find an orthogonal basis for the 3-dimensional subspace V of 4 that consists of all solutions of the equation x1 + x2 + x3 − x4 = 0.

A norm on V provides a definition of the distance d( x, y) between any two points x and y of V: Note that a distance function d defined in this way satisfies the following three conditions: for any three points x, y, z. Conditions D1 and D2 follow immediately from N1 and N2, respectively, while by N3. 1 indicates why N3 (or D3) is referred to as the triangle inequality. 1 The distance function that comes in this way from the Euclidean norm is the familiar Euclidean distance function Thus far we have seen that an inner product on the vector space V yields a norm on V, which in turn yields a distance function on V, except that we have not yet verified that the norm associated with a given inner product does indeed satisfy the triangle inequality.

Any set of n linearly independent vectors in an n-dimensional vector space has this property. 4 If the vectors v1, . . , vn in the n-dimensional vector space V are linearly independent, then they constitute a basis for V, and furthermore generate V uniquely. PROOF Given , the vectors v, v1, . . 1 there exist numbers x, x1, . . , xn, not all zero, such that If x = 0, then the fact that v1, . . , vn are linearly independent implies that x1 = · · · = xn = 0. Therefore x ≠ 0, so we solve for v: Thus the vectors v1, .