HARVARD MATH 21B  GRAM SCHMIDT AND QR (2 pages)
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GRAM SCHMIDT AND QR
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 Math 21b  Linear Algebra and Differential Equations
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GRAM SCHMIDT AND QR Math 21b O Knill GRAM SCHMIDT PROCESS Let v1 vn be a basis in V Let u1 v1 and w 1 u1 u1 The Gram Schmidt process recursively constructs from the already constructed orthonormal set w 1 w i 1 which spans a linear space Vi 1 the new vector ui vi projVi 1 vi which is orthogonal to Vi 1 and then normalizing ui to to get w i ui ui Each vector w i is orthonormal to the linear space Vi 1 The vectors w 1 w n form an orthonormal basis in V EXAMPLE 1 1 2 Find an orthonormal basis for v1 0 v2 3 and v3 2 5 0 0 SOLUTION 1 1 w 1 v1 v1 0 0 0 0 2 u2 u2 1 1 v2 w 1 3 w 2 u2 v2 projV1 v2 v2 w 0 0 0 0 3 u3 v3 projV2 v3 v3 w 1 v3 w 1 w 2 v3 w 2 0 w 3 u3 u3 0 5 1 QR FACTORIZATION The formulas can be written as v1 v1 w 1 r11 w 1 vi w 1 vi w 1 w i 1 vi w i 1 ui w i ri1 w 1 rii w i vn w 1 vn w 1 w n 1 vn w n 1 un w n rn1 w 1 rnn w n which means in matrix form 1 A v1 vm w r11 w m 0 0 where A and Q are n m matrices and R is a m m matrix r12 r22 0 r1m r2m QR rmm Any matrix A with linearly independent columns can be decomposed as A QR where Q has orthonormal column vectors and where R is an upper triangular square matrix BACK TO THE EXAMPLE The matrix with the vectors v1 v2 v3 v1 v1 w 1 v2 w 1 v2 w 1 u2 w 2 v3 w 1 v3 w 1 w 2 v3 w 2 u3 w 3 1 so that Q 0 0 2 1 1 is A 0 3 2 0 0 5 0 0 2 1 1 0 and R 0 3 0 1 0 0 1 2 5 PRO MEMORIA While building the matrix R we keep track of the vectors u i during the Gram Schmidt procedure At the end you have vectors ui vi w i and the matrix R has ui in the diagonal as well as the dot products w i vj in the upper right triangle PROBLEM Make the QR decomposition of A 0 1 1 1 w 2 u2 A QR 1 0 0 1 0 1 1 1 w 1 0 1 u2 1 1 0 1 1 0 WHY do we care to have an orthonormal basis An orthonormal basis looks like the standard basis v 1 1 0 0 vn 0 0 1 Actually we will see that an orthonormal basis into a standard basis or a
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