Lecture 4 - 4 4.1 QR Factorization Reduced vs Full QR...

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4 QR Factorization 4.1 Reduced vs. Full QR Consider A C m × n with m n . The reduced QR factorization of A is of the form A = ˆ Q ˆ R, where ˆ Q C m × n with orthonormal columns and ˆ R C n × n an upper triangular matrix such that ˆ R ( j, j ) = 0, j = 1 , . . . , n . As with the SVD ˆ Q provides an orthonormal basis for range( A ), i.e., the columns of A are linear combinations of the columns of ˆ Q . In fact, we have range( A ) = range( ˆ Q ). This is true since A x = ˆ Q ˆ R x = ˆ Q y for some y so that range( A ) range( ˆ Q ). Moreover, range( Q ) range( ˆ A ) since we can write A ˆ R - 1 = ˆ Q because ˆ R is upper triangular with nonzero diagonal elements. (Now we have ˆ Q x = A ˆ R - 1 x = A y for some y .) Note that any partial set of columns satisfy the same property, i.e., span { a 1 , . . . , a j } = span { q 1 , . . . , q j } , j = 1 , . . . , n. In order to obtain the full QR factorization we proceed as with the SVD and extend ˆ Q to a unitary matrix Q . Then A = QR with unitary Q C m × m and upper triangular R C m × n . Note that (since m n ) the last m - n rows of R will be zero. 4.2 QR Factorization via Gram-Schmidt We start by formally writing down the QR factorization A = QR as a 1 = q 1 r 11 = q 1 = a 1 r 11 (14) a 2 = q 1 r 12 + q 2 r 22 = q 2 = a 2 - r 12 q 1 r 22 (15) . . . . . . (16) a n = q 1 r 1 n + q 2 r 2 n + . . . + q n r nn = q n = a n - n i =1 r in q i r nn (17) Note that in these formulas the columns a j of A are given and we want to determine the columns q j of Q and entries r ij of R such that Q is orthonormal, i.e., q * i q j = δ ij , (18) R is upper triangular and A = QR . The latter two conditions are already reflected in the formulas above. Using (14) in the orthogonality condition (18) we get q * 1 q 1 = a * 1 a 1 r 2 11 = 1 so that r 11 = a * 1 a 1 = a 1 2 . 36

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Note that we arbitrarily chose the positive square root here (so that the factorization becomes unique). Next, the orthogonality condition (18) gives us q * 1 q 2 = 0 q * 2 q 2 = 1 . Now we apply (15) to the first of these two conditions. Then q * 1 q 2 = q * 1 a 2 - r 12 q * 1 q 1 r 22 = 0 . Since we ensured q * 1 q 1 = 1 in the previous step, the numerator yields r 12 = q * 1 a 2 so that q 2 = a 2 - ( q * 1 a 2 ) q 1 r 22 . To find r 22 we normalize, i.e., demand that q * 2 q 2 = 1 or equivalently q 2 2 = 1. This immediately gives r 22 = a 2 - ( q * 1 a 2 ) q 1 2 . To fully understand how the algorithm proceeds we add one more step (for n = 3). Now we have three orthogonality conditions: q * 1 q 3 = 0 q * 2 q 3 = 0 q * 3 q 3 = 1 . The first of these conditions together with (17) for n = 3 yields q * 1 q 3 = q * 1 a 3 - r 13 q * 1 q 1 - r 23 q * 1 q 2 r 33 = 0 so that r 13 = q * 1 a 3 due to the orthonormality of columns q 1 and q 2 . Similarly, the second orthogonality condition together with (17) for n = 3 yields q * 2 q 3 = q * 2 a 3 - r 13 q * 2 q 1 - r 23 q * 2 q 2 r 33 = 0 so that r 23 = q * 2 a 3 . Together this gives us q 3 = a 3 - ( q * 1 a 3 ) q 1 - ( q * 2 a 3 ) q 2 r 33 and the last unknown, r 33 , is determined by normalization, i.e., r 33 = a 3 - ( q * 1 a 3 ) q 1 - ( q * 2 a 3 ) q 2 2 .
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