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Highlights Math 304 Linear Algebra From last time: I Harold P. Boas inner products and norms Today: I [email protected] orthonormal sets June 22, 2006 Example in R 3 √1 2 Example continued Let u1 = 0 , u2 = − √1 2 √1 3 − √1 , 3 1 √ 3 √1 6 √2 6 . √1 6 and u3 = These vectors form an orthonormal set: each vector is orthogonal to the others, and each vector has norm equal to 1. The matrix U whose columns are u1 , u2 , and u3 is called an orthogonal matrix. Notice that U T U = I (the identity matrix), so U T = U −1 . This property characterizes orthogonal matrices. An orthogonal matrix preserves the scalar product: namely, hUx, Uyi = hx, yi. Here is why: hUx, Uyi = (Ux)T Uy = xT U T Uy = xT y = hx, yi. Therefore an orthogonal matrix also preserves the norm: kUxk = kxk. √1 2 Let y = 2 0 − 1 −√ 2 √1 3 3 − √13 √1 3 √1 6 4 √26 √1 6 + = 2u1 − 3u2 + 4u3 . Problem. Determine kyk. 2 Solution. Since y = U −3, and since multiplication by an 4 orthogonal matrix preserves the norm, 2 √ √ −3 = 4 + 9 + 16 = 29. kyk = 4 In general, kc1 u1 + c2 u2 + c3 u3 k2 = c12 + c22 + c32 . The analogous statement for an arbitrary orthonormal set is Parseval’s formula. More on the example: Fourier coefficients Projection and approximation (1, 2, 3)T Problem. Express the vector v = as a linear combination of u1 , u2 , and u3 . In other words, find coefficients c1 , c2 , and c3 such that v = c1 u1 + c2 u2 + c3 u3 or 1 1 1 √ √ √ 1 3 6 2 1 2 = c1 0 + c2 − √ + c3 √2 . 3 6 3 √1 √1 − √1 2 3 6 Solution. No row reduction necessary! Take the inner product + c2 u2 + c3 u3 , u1 i = c1 by with u1 : thus hv, u1 i = hc1 u1√ orthonormality, so√c1 = −2/ 2 . Similarly, √ c2 = hv, u2 i = 2/ 3 , and c3 = hv, u3 i = 8/ 6. In general, any vector v can be represented in terms of an orthonormal basis u1 , u2 , and u3 via v = hv, u1 iu1 + hv, u2 iu2 + hv, u3 iu3 . Problem. Find the projection of the vector v = (2, 1, 1)T onto the plane spanned by u1 and u2 . Solution. We can write v = hv, u1 iu1 + hv, u2 iu2 + hv, u3 iu3 , but u3 is orthogonal to the plane, so the projection equals 1 2 7 2 3 6 hv, u1 iu1 + hv, u2 iu2 = 0 + − 23 = − 23 . 1 2 − 12 6 3