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The example at left is that of an orthogonal array with symbol set {1,2} and strength 2. Notice that the four ordered pairs (2-tuples) formed by the rows restricted to the first and third columns, namely (1,1), (2,1), (1,2) and (2,2), are all the possible ordered pairs of the two element set and each appears exactly once.
And second, for each row or column of the matrix, there is a constraint that the sum of the cells in that row or column equal one. The row and column constraints define a linear subspace of dimension n 2 − 2n + 1 in which the Birkhoff polytope lies, and the non-negativity constraints define facets of the Birkhoff polytope within that subspace.
A set equipped with a total order is a totally ordered set; [5] the terms simply ordered set, [2] linearly ordered set, [3] [5] toset [6] and loset [7] [8] are also used. The term chain is sometimes defined as a synonym of totally ordered set , [ 5 ] but generally refers to a totally ordered subset of a given partially ordered set.
The ordered pair (a, b) is different from the ordered pair (b, a), unless a = b. In contrast, the unordered pair, denoted {a, b}, always equals the unordered pair {b, a}. Ordered pairs are also called 2-tuples, or sequences (sometimes, lists in a computer science context) of length 2. Ordered pairs of scalars are sometimes called 2-dimensional ...
In mathematics a linear inequality is an inequality which involves a linear function. A linear inequality contains one of the symbols of inequality: [1] < less than > greater than; ≤ less than or equal to; ≥ greater than or equal to; ≠ not equal to; A linear inequality looks exactly like a linear equation, with the inequality sign ...
In linear algebra, Weyl's inequality is a theorem about the changes to eigenvalues of an Hermitian matrix that is perturbed. It can be used to estimate the eigenvalues of a perturbed Hermitian matrix.
For instance, to solve the inequality 4x < 2x + 1 ≤ 3x + 2, it is not possible to isolate x in any one part of the inequality through addition or subtraction. Instead, the inequalities must be solved independently, yielding x < 1 / 2 and x ≥ −1 respectively, which can be combined into the final solution −1 ≤ x < 1 / 2 .
There exist y 1, y 2 such that 6y 1 + 3y 2 ≥ 0, 4y 1 ≥ 0, and b 1 y 1 + b 2 y 2 < 0. Here is a proof of the lemma in this special case: If b 2 ≥ 0 and b 1 − 2b 2 ≥ 0, then option 1 is true, since the solution of the linear equations is = and =.