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Lecture

# Gram-Schmidt method; QR-decomposition.pdf

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School
Department
Statistics
Course
STAT312
Professor
Douglas Wiens
Semester
Fall

Description
34 6. Gram-Schmidt method; QR-decomposition  The trace of a square matrix is the sum of its diagonal elements. A useful identity is (how?) (AB ) = (BA ) Thus products within traces can be rearranged cyclically: (ABC ) = (CAB ) = (BCA ) but not necessari=y (ACB ) It will be shown that for an idempotent matrix, = . A consequence is that ³ ´ dim X ) = I H) = (I H ) = (H) µ ³ 0 ´ 1 0 = X X X X µ ³ ´ ¶ = X X X X0 1 = Similarly(X ) = (H ), (H) = . 35  A matrixQ × is orthogonal if the columns are mutually orthogonal, and have unit norm. Equiv- alently (why?) QQ 0= Q Q = I IfQ is orthogonal theQykk = k yk for any × 1 vectoy  norms are preserved. Simi- larly, angles between vectors are also preserved (why?). Geometrically, an orthogonal transfor- mation is a rigid motion  it corresponds to a rotation and/or an interchange of two or more axes. Rotation through an anin the plane: Ã ! cos sin Q = sin cos Interchange of axes in the plane: Ã ! 0 1 Q = 1 0 36  Gram-Schmidt Theorem: Every -dimensional vector space has an orthogonal basis. Proof: Start with any basv1 v . Normal- izev1 to get a unit vector (i.e. a vector with unit norm) q ; in general suppose that mutually 1 orthogonal unit vectoq1 q have been con- structed, wiqh a linear combinationvof v . 1 Dene X 0 H = q q ³1 ´ I H v q = °³ ´ +1 ° +1 ° I H v ° +1 For instanceq2= ... . Then: (i) the denominator is non-zero (why?); (ii)q +1 is a linear combination v1 v +1 ; (iiiq +1 q1 q (why?). Continuing this process results in mutually orthog
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