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Example text

The problem of solving Bx = c is indeed ill-conditioned, but the algorithm (Gaussian elimination without pivoting) does not introduce any unnecessary loss of accuracy. , boundary value problems for ordinary and partial differential equations, matrices arise where a large proportion of the elements are equal to zero. If the nonzero elements are concentrated around the main diagonal, then the matrix is called a band matrix. More precisely, a matrix A is said to be a band matrix if there are natural numbers p and q, such that aij = 0 if j − i > p or i − j > q.

The LU decomposition of the same matrix with permuted rows is 1 1 ǫ 1 = 1 0 1 1 ǫ 1 0 1−ǫ . No growth of matrix elements occurs, and the factors are very well-conditioned: κ∞ (L) ≈ 1, κ∞ (U ) ≈ 4. 8 The matrix A= 2 1 1 , 2 is symmetric and positive definite. It is well-conditioned: κ∞ (A) = 3. The LDLT decomposition is 1 0 2 0 1 12 A= 1 . 1 0 32 0 1 2 No growth of matrix elements occurs. Next, consider the matrix B= ǫ 1 1 1/ǫ + 1 . ‘‘boo 2005/ page 28 Chapter 3. Linear Systems and Least Squares B is also symmetric and positive definite.

1 (SVD). 1) where U ∈ Rm×m and V ∈ Rn×n are orthogonal, and Σ ∈ Rn×n is diagonal Σ = diag(σ1 , σ2 , . . , σn ), σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0. Proof. The assumption m ≥ n is no restriction: in the other case, just apply the theorem to AT . We give a proof along the lines of that in . Consider the maximization problem sup Ax 2 . x 2 =1 Since we are seeking the supremum of a continuous function over a closed set, the supremum is attained for some vector x.