[SOLVED] MA514 -HW 5

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Exercise 20.2

We can say that L is lower triangular and also l_ij = 0 for i−j > p. We can say that U is upper triangular and also that u_ij = 0 for j − i > p. To see this, note that since A satisfies the conditions of exercise 20.1, A has an LU factorization A = LU. When i − j > p, because U is upper triangular we have

j a_ij = ^Xl_iku_kj = 0      for i = 1,…,m − p − 1 .

k=1

Therefore, l_ij = 0 for i−j > p. Because L is lower triangular, when j−i > p we have

i a_ij = ^Xl_iku_kj = 0,     for i = 1,…,m − p − 1 .

k=1

Therefore, u_ij = 0 for j − i > p.

Exercise 23.1

Let A be a nonsingular square matrix and let A = QR and A^∗A = U^∗U be

QR and Cholesky factorizations, respectively, with the usual normalizations r_jj,u_jj > 0. Then it is true that R = U. To see this, first we note that since A is nonsingular, A has a unique QR factorization and A^∗A = (QR)^∗QR =

R^∗R.               Also, because A is nonsingular, Ax 6= 0 for every x 6= 0.            Then

0, which shows that A^∗A is positive definite. And

because (A^∗A)^∗ = A^∗(A^∗)^∗ = A^∗A, A^∗A is also hermitian. So by Theorem 23.1, the Cholesky factorization A^∗A = U^∗U of A^∗A is unique. But then

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A^∗A = R^∗R = U^∗U and both factorizations must be unique, we conclude that R = U.

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