This document discusses describing the set S = {x E R^n : ai^Tx = alphai,i = 1,... ,m} for different values of n and m. It explains that for n = m, the set S is the solution to the system of linear equations Ax = b, where A is the matrix with rows ai and b is the vector with elements alphai. When m < n, there may be an infinite number of solutions or no solution, depending on the consistency of the underdetermined system of equations. When m > n, the set S could be empty, finite, or infinite depending on the linear dependence and consistency of the overdetermined system.

1. Linear Optimization Linear Optimization Problem 3. For i = 1,. .. , m, let ai = (ai1,. . . , ain)T in
R^n and alphai in R. a) For n = 1,2,3 and n > 3 with m = n describe the set S = {x E R^n : ai^Tx
= alphai,i = 1,... ,m} b) How does this description change when m n? d) Express the set S in
matrix notation.
Solution
I = 1,2,3,….M A1= A2= A3= AM= A11 A21 A31 …………….. AM1
A12 A22 A32 …………….. AM2 A13 A23 A33 …………….. AM3 ……………..
…………….. …………….. …………….. …………….. A1N A2N A3N ……………..
AMN USING ' TO INDICATE TRANSPOSE ..THAT IS TRANSPOSE OF Ai = Ai'
USING B INSTEAD OF ALPHA FOR EASE IN TYPING ..SO [ALPHA]I = Bi S = [ X IS
AN ELEMENT OF R^N :Ai' *X = Bi ] a)……. IN GENERAL IN THIS CASE WE
SHALL HAVE N LINEAR EQNS.IN N VARIABLES ..X1,X2,…XN ..SEE BELOW …
DEPENDING ON THE LINEAR DEPENDENC & CONSISTENCY CHARACTER OF THE
EQNS. WE MAY GET S AS A NULL SET OR FINITE SET OR AN INFINITE SET
FOR N= 1 = M …... WE GET THE SET BY THE EQN... … A1= A11 * X1 =
B1 X1 = B1 /[A11] ..SO … S = [ X1] = [B1/A11] FOR N= 2 = M …...
WE GET THE SET BY THE EQN... … A = X= B= A1= A2= A11 A21 * X1
= B1 A12 A22 X2 B2 SO WE GET ….. S =[ X ] = [X1,X2] …GIVEN BY ….
X= A INVERSE * B FOR N= 3 = M …... WE GET THE SET BY THE EQN... …
A = X= B= A1= A2= A3= A11 A21 A31 X1 B1 A12 A22 A32 * X2 =
B2 A13 A23 A33 X3 B3 SO WE GET ….. S =[ X ] = [X1,X2,X3] …GIVEN BY ….
X= A INVERSE * B IN GENERAL FOR N=M WE GET THE SET BY EQN…
A= X = B = A1= A2= A3= AM= A11 A21 A31 …………….. AN1 X1 B1 A12
A22 A32 …………….. AN2 * X2 = B2 A13 A23 A33 …………….. AN3 X3 B3
…………….. …………….. …………….. …………….. …………….. ……………..
…………….. A1N A2N A3N …………….. ANN XN = BN SO WE GET ….. S =[ X ]
= [X1,X2,X3…..XN] …GIVEN BY …. X= A INVERSE * B b)…… M < N
IN GENERAL IN THIS CASE WE SHALL HAVE M LINEAR EQNS.IN N VARIABLES
..X1,X2,…XN .. SINCE THE NUMBER OF EQNS. ARE LESS THAN THE NUMBER
OF VARIABLES , WE MAY GET S AS A NULL SET OR AN INFINITE SET
DEPENDING ON THE CONSISTENCY OF THE EQNS c)….. M > N IN
GENERAL IN THIS CASE WE SHALL HAVE M LINEAR EQNS.IN N VARIABLES
..X1,X2,…XN .. SINCE THE NUMBER OF EQNS. ARE MORE THAN THE NUMBER
OF VARIABLES , DEPENDING ON THE LINEAR DEPENDENC & CONSISTENCY

2. CHARACTER OF THE EQNS. WE MAY GET S AS A NULL SET OR FINITE SET OR
AN INFINITE SET d)…. ALREADY SHOWN ABOVE ..