By theorem 3 an orthonormal set is necessarily

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By Theorem 3 an orthonormal set is necessarily linearly independent. DEFINITION Orthonormal Basis A basis for an inner product space V which is an orthonormal set is called an orthonormal basis of V . EXAMPLE 14 The standard basis of R n is an orthonormal basis under the standard inner product. EXAMPLE 15 The standard basis for R 3 is not an orthonormal basis under the inner product ( vectorx, vector y ) = x 1 y 1 + 2 x 2 y 2 + x 3 y 3 since vectore 2 = 0 1 0 is not a unit vector under this inner product.
Section 3.2 Orthogonality and Length 11 EXAMPLE 16 Let B = 1 3 1 1 - 1 , 1 2 1 0 1 , 1 6 1 - 2 - 1 . Verify that B is orthonormal in R 3 . Solution: We first verify that each vector is a unit vector. We have (Bigg 1 3 1 1 - 1 , 1 3 1 1 - 1 )Bigg = 1 3 + 1 3 + 1 3 = 1 (Bigg 1 2 1 0 1 , 1 2 1 0 1 )Bigg = 1 2 + 0 + 1 2 = 1 (Bigg 1 6 1 - 2 - 1 , 1 6 1 - 2 - 1 )Bigg = 1 6 + 4 6 + 1 6 = 1 We now verify that the set is orthogonal. (Bigg 1 3 1 1 - 1 , 1 2 1 0 1 )Bigg = 1 6 (1 + 0 - 1) = 0 (Bigg 1 3 1 1 - 1 , 1 6 1 - 2 - 1 )Bigg = 1 18 (1 - 2 + 1) = 0 (Bigg 1 2 1 0 1 , 1 6 1 - 2 - 1 )Bigg = 1 12 (1 + 0 - 1) = 0 Thus, B is an orthonormal set in R 3 .
12 Chapter 3 Inner Products EXAMPLE 17 Turn { 1 , x, 3 x 2 - 2 } into an orthonormal basis for P 2 under the inner product ( p, q ) = p ( - 1) q ( - 1) + p (0) q (0) + p (1) q (1). Solution: In Example 9 we showed that { 1 , x, 3 x 2 - 2 } is orthogonal. Hence, it is a linearly independent set of three vectors in P 2 and thus is a basis for P 2 . To turn it into an orthonormal basis, we just need to normalize each vector. We have bardbl 1 bardbl = 1 2 + 1 2 + 1 2 = 3 bardbl x bardbl = radicalbig ( - 1) 2 + 0 2 + 1 2 = 2 bardbl 3 x 2 - 2 bardbl = radicalbig 1 2 + ( - 2) 2 + 1 2 = 6 Hence, an orthonormal basis for P 2 is braceleftbigg 1 3 , x 2 , 3 x 2 - 2 6 bracerightbigg EXAMPLE 18 Show that the set { 1 , sin x, cos x } is an orthogonal set in C [ - π, π ] under ( f, g ) = integraldisplay π - π f ( x ) g ( x ) dx , then make it an orthonormal set. Solution: ( 1 , sin x ) = integraldisplay π - π 1 · sin x dx = 0 ( 1 , cos x ) = integraldisplay π - π 1 · cos x dx = 0 ( sin x, cos x ) = integraldisplay π - π sin x · cos x dx = 0 Thus, the set is orthogonal. Next we find that ( 1 , 1 ) = integraldisplay π - π 1 2 dx = 2 π ( sin x, sin x ) = integraldisplay π - π sin 2 x dx = π ( cos x, cos x ) = integraldisplay π - π cos 2 x dx = π Hence, bardbl 1 bardbl = 2 π , bardbl sin x bardbl = π , and bardbl cos x bardbl = π . Therefore, the set braceleftbigg 1 2 π , sin x π , cos x π bracerightbigg is orthonormal THEOREM 5 If vectorv is any vector in an inner product space V with inner product ( , ) and B = { vectorv 1 , . . . ,vectorv n } is an orthonormal basis for V , then vectorv = ( vectorv,vectorv 1 ) vectorv 1 + · · · + ( vectorv,vectorv n ) vectorv n
Section 3.2 Orthogonality and Length 13 Proof: By Theorem 4 we have vectorv = ( vectorv,vectorv 1 ) bardbl vectorv 1 bardbl 2 vectorv 1 + · · · + ( vectorv,vectorv n ) bardbl vectorv n bardbl

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