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Lecture 37

MATH125 Lecture 37: 37

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University of Alberta

Final Lecture 37, April 12, 2017. Theorem. Let Q be an orthogonal matrix. a. The matrix Q ▯1 is orthogonal. b. detQ = ▯1. c. If ▯ is an eigenvalue of Q, then ▯ = ▯1. d. If Q1and Q ar2 orthogonal n ▯ n matrices, then so is the product Q Q . 1 2 ▯1 T Proof. (a) Q = Q is orthogonal. (b) 1 = detI = det(Q Q) = detQ detQ = (detQ) , so that detQ = ▯1. (c) If v is an eigenvector with eigenvalue ▯, then jjvjj = jjQvjj = jj▯vjj = j▯jjjvjj so that j▯j = 1. (d) (Q1Q 2 Q Q1=2Q Q Q2Q 1 Q1I2 = I.2 2 ▯ x5.2: Orthogonal Complements n De▯nition. Let W be a subspace in R . We say that a vector v in R is orthogonal to W, if v is orthogonal to every vector in W. The set of all vectors that are orthogonal to W is called the orthogonal complement of W, denoted W . That is, W ? = fv in Rn : v ▯ w = 0 for all w in Wg: Theorem. Let W be a subspace of R . n a. W ? is a subspace of R . ? ? b. (W ) = W. c. W \ W ? = f0g. ? d. If W is spanned by vectors w1;:::;wk, then v is in W is and only if v ▯iw = 0 for all i = 1;:::;k. Proof. (a) 0 is in W because 0▯w = 0 for any w in W. If u and v are ? in W , then for any w in W we have (u + v) ▯ w = u ▯ w + v ▯ w = 0 + 0 = 0 ? so that u+v is in W, too. Finally, if u is W , then for any scalar c we have
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