Application of LU-factorizationComputing determinantsDetermening Positive definitenessStability of LUPivotingApplication of LU-factorizationStability of LULecture 17: LU-factorization, continuedAmos RonUniversity of Wisconsin - MadisonMarch 19, 2021Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUOutline1Application of LU-factorizationComputing determinantsDetermening Positive definiteness2Stability of LUPivotingAmos Ron CS513, remote learning S21Application of LU-factorizationStability of LUBlank pageAmos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessOutline1Application of LU-factorizationComputing determinantsDetermening Positive definiteness2Stability of LUPivotingAmos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessDeterminant of triangular matrices, and generalmatricesAssume A is triangular (upper, lower), then:The eigenvalues of A are the diagonal entries of A.det(A) =Qmi=1A(i, i).Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessDeterminant of triangular matrices, and generalmatricesAssume A is triangular (upper, lower), then:The eigenvalues of A are the diagonal entries of A.det(A) =Qmi=1A(i, i).Computing det(A) for a general square matrixOption 1: Directly from the definition. Complexity: O(m!) -hopeless.Option 2: Using the multiplication theorem:det(BC) = det(B) det(C).Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessDeterminant of triangular matrices, and generalmatricesComputing det(A) for a general square matrixOption 1: Directly from the definition. Complexity: O(m!) -hopeless.Option 2: Using the multiplication theorem:det(BC) = det(B) det(C).Algorithm for computing det(A)LU-factor A.det(A) =Qmi=1U(i, i) (since det(L) = 1).Complexity: O(m3).Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessThe three characterizations of positive definitenessTheorem: Let A be square symmetric, invertible, with LDUfactorization. TFCAE:A is SPD.σ(A) > 0.All the main principal minor are positive.The diagonal entries of D in A = LDU are all positive.Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessThe three characterizations of positive definitenessTheorem: Let A be square symmetric, invertible, with LDUfactorization. TFCAE:A is SPD.σ(A) > 0.All the main principal minor are positive.The diagonal entries of D in A = LDU are all positive.Proof: We only prove that the last condition is equivalent to therest.Since A is symmetric,A = U0DU.Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessThe three characterizations of positive definitenessProof: We only prove that the last condition is equivalent to therest.Since A is symmetric,A = U0DU.So, let v 6= 0.(Av, v) = ((U0DU)v, v) = (DUv, Uv) = (Dw, w).With w = Uv. Since U is intertible, and v 6= 0, we have w 6= 0.If D(i, i) > 0, ∀i, then (Dw, w) > 0 (easy, and we already arguedthat before). Then A is SPD.If D(i, i) ≤ 0 for some i, then we can choose w := ei(i.e., wechoose v := U−1ei).Then (Dw, w) = ...Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessThe three characterizations of positive definitenessProof: We only prove that the last condition is equivalent to therest.Since A is symmetric,A = U0DU.So, let v 6= 0.(Av, v) = ((U0DU)v, v) = (DUv, Uv) = (Dw, w).With w = Uv. Since U is intertible, and v 6= 0, we have w 6= 0.If D(i, i) > 0, ∀i, then (Dw, w) > 0 (easy, and we already arguedthat before). Then A is SPD.If D(i, i) ≤ 0 for some i, then we can choose w := ei(i.e., wechoose v := U−1ei).Then (Dei, ei) = D(i, i) ≤ 0, and therefore A is not SPD.Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUComputing determinantsDetermening Positive definitenessBlank pageAmos Ron CS513, remote learning S21Application of LU-factorizationStability of LUPivotingOutline1Application of LU-factorizationComputing determinantsDetermening Positive definiteness2Stability of LUPivotingAmos Ron CS513, remote learning S21Application of LU-factorizationStability of LUPivotingA 2 × 2 examplesConsider the matrix A −11 0where  is small.When  = 0, A is orthogonal, hence cond(A) = 1. Therefore, forsmall , we havecond(A) ≈ 1.Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUPivotingA 2 × 2 examplesConsider the matrix A −11 0where  is small.When  = 0, A is orthogonal, hence cond(A) = 1. Therefore, forsmall , we havecond(A) ≈ 1.However, the LU-factorization of A isL =1 0−11U = −10 −1Then:cond(L) ≈ −1, cond(U) ≈ −2.So, the factorization is unstableAmos Ron CS513, remote learning S21Application of LU-factorizationStability of LUPivotingA 2 × 2 examplesWhat to do?The problem is that in the algorithm we definev1(2 : m) = A(2 : m, 1)/A(1, 1),and the pivot A(1, 1) is small compared to other entries of A,hence v1contains large entries. This automatically makes Lill-conditioned.Key: Do not use pivots that are small relative to other entries.Amos Ron CS513, remote learning S21Application of LU-factorizationStability of LUPivotingA 2 × 2 examplesKey: Do not use pivots that are small relative to other entries.Solution: Assuming that we are allowed to shuffle the order ofthe rows/columns we do so.Shuffling both rows and columns: full pivotingShuffling only rows: partial pivotingAmos Ron CS513, remote learning S21Application of LU-factorizationStability of LUPivotingHow to perform the pivoting?At each step: The input matrix Aj−1has already undergonesome pivoting, i.e., some row and columns were shuffled.In the j0th, step, we need to create the vectorvj(j : m) = Aj−1(j : m, j)/Aj−1(j, j).So, we want to bring to the (j, j)-location a large entry.We can choose a row among the j, . . . m rows of Aj−1.And, if we do full pivoting, we can choose a column from thej, . . . , m columns of Aj−1Amos Ron CS513, remote learning