Solving a system of equations 1 x x x 3 = 3 2 x x x 3 = 2 3 x x x 3 = 3

Solving a system of equations 1 x x x 3 = 3 2 x x x 3 = 2 3 x x x 3 = = 2 no solution

Solving a system of equations a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 A x = b has a solution iff rank (A) = rank (A b)

Solving a system of equations A x = b has a solution iff rank (A) = rank (A b) rank (A) < rank (A b) then  y such that y T A = 0 and y T b  0

Solving a system of equations A x = b has a solution iff rank (A) = rank (A b) rank (A) < rank (A b) then  y such that y T A = 0 and y T b  0 y is a witness that A x = b does not have a solution

Solving a system of equations y T A = 0 and y T b  0 y is a witness that A x = b does not have a solution A x = b  y T (A x) = y T b  (y T A) x = y T b  0 = y T b  a contradiction

Solving a system of equations Theorem: If Ax=b doesn’t have solution then  y such that y T A =0 and y T b  0

Solving a system of equations 1 x x x 3 = 3 2 x x x 3 = 2 3 x x x 3 = 1

Solving a system of equations 1 x x x 3 = 3 2 x x x 3 = 2 3 x x x 3 = 1 x 1 = -3 x 2 = 2 x 3 = 0

Back to linear programming 1 x x x 3 = 3 2 x x x 3 = 2 3 x x x 3 = 1 x 1  0 x 2  0 x 3  0 I.e., we want a non-negative solution.

Back to linear programming 1 x x x 3 = 3 2 x x x 3 = 2 3 x x x 3 = 1 x 1  0 x 2  0 x 3  0 I.e., we want a non-negative solution. 0 1 x 1 + x 2 + x 3 = -1

y T A  0 and y T b < 0 y is a witness that A x = b, x  0 does not have a solution A x = b  y T (A x) = y T b  (y T A) x = y T b  non-negative = negative, a contradiction Back to linear programming

Theorem (Farkas): If Ax=b, x  0 doesn’t have a solution then  y such that y T A  0 and y T b < 0 Theorem: If Ax=b doesn’t have solution then  y such that y T A =0 and y T b  0 Back to linear programming

Theorem (Farkas): If Ax=b, x  0 doesn’t have a solution then  y such that y T A  0 and y T b < 0 Idea of the proof Ax, x  0 b

Theorem (Farkas): If Ax=b, x  0 doesn’t have a solution then  y such that y T A  0 and y T b < 0 Idea of the proof Ax, x  0 b cTxcTx separating hyperplane S=convex, b not in S  c such that (  x  S)c T x > c T b

Theorem (Farkas): If Ax=b, x  0 doesn’t have a solution   y such that y T A  0 and y T b < 0 Duality max c T x Ax=b x  0 min y T b y T A  c T

a 1 x a n x n  b a 1 x a n x n  b + y, y  0 a 1 x a n x n – y  b, y  0  “  ”   “=” and non-negativity

a 1 x a n x n  b a 1 x a n x n  b a 1 x a n x n  b  “  ”   “  ” a 1 x a n x n  b -a 1 x a n x n  -b

Duality max c 1 T x 1 +c 2 T x 2 + c 3 T x 3 + c  T x 4 A 1 x 1 =b 1 A 2 x 2  b 2 A 3 x 3 = b 3 A 4 x 4  b 4 x 1  0,x 2  0 min y 1 T b 1 +y 2 T b 2 +y 3 T b 3 + y 4 T b 4 y 1 T A 1  c 1 T y 2 T A 2  c 2 T y 3 T A 3 = c 3 T y 4 T A 4 = c 4 T y 2  0, y 4  0

Solving linear programs Simplex (Danzig, 40’s) Ellipsoid (Khachiyan, 80’s) Interior point (Karmakar, 80’s)