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Q8SE

Expert-verifiedFound in: Page 395

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Question 8: Use Exercise 7 to show that if A is positive definite, then A has a LU factorization, \(A = LU\), where U has positive pivots on its diagonal. (The converse is true, too).**

It is proved that *A* has a LU factorization \(A = LU\).

**Theorem 12 in section 6.4 **states that when \(A\) is a \(m \times n\) matrix that is **linearly independent columns**, then \(A\) may be factored as \(A = QR\), with \(Q\) is a \(m \times n\) matrix wherein columns provide an **orthonormal basis **for \({\mathop{\rm Col}\nolimits} A\), and \(R\) is an \(n \times n\) upper triangular invertible matrix which has positive entries on its diagonal.

Assume that \(A\) is positive definite, and suppose that Cholesky factorization of \(A = {R^T}R\) where \(R\) is an upper triangular and having positive diagonal entries. Consider that \(D\) as the diagonal matrix with diagonal entries equal to diagonal entries of \(R\). The matrix \(L = {R^T}{D^{ - 1}}\) is lower triangular wherein 1’s on its diagonal because right-multiply by a diagonal matrix scales the columns of the matrix on its left.

When \(U = DR\) then \(A = {R^T}{D^{ - 1}}DR = LU\).

Conversely, consider that *A* contains a LU factorization, that is \(A = LU\), where \(U\) contains positive pivots on its diagonal. Consider \(D\) as the diagonal matrix and \(\sqrt {{u_{11}}} , \ldots ,\sqrt {{u_{nn}}} \) on its diagonal. The matrix \(V = {\left( {{D^2}} \right)^{ - 1}}U\) is upper triangular wherein 1’s on its diagonal because multiply by a diagonal matrix on its left scales the rows of the matrix on its right and we obtain \(A = L{D^2}V\). As \(A\) is symmetric, then \({A^T} = {\left( {L{D^2}V} \right)^T} = {V^T}{D^2}{L^T} = A\), and \(L = {V^T}\). When \(R = DV = {D^{ - 1}}U\) then \(A = {V^T}DDV = {\left( {DV} \right)^T}\left( {DV} \right) = {R^T}R\).

Hence, it is proved that *A* has a LU factorization \(A = LU\).

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