A is a 4 times 5 rank 2 matrix for which the vectors X1 17

A is a 4 times 5, rank 2, matrix for which the vectors X_1 = [17, 6, -13, 1, 2] X_2 = [1, 2, 3, 4, 5]^t, and X_3 = [5, 4, 3, 2, 1]^t all satisfy AX = 0. Prove that X_1, X_2, and X_3 span the solution space to the system AW = 0 Is it possible to find a 4 times 5, rank 3 matrix B such that the given X_i al satisfy BX_i = 0?

Solution

(a). The rank–nullity theorem states that the sum of the rank and the nullity of a matrix is equal to the number of columns of the matrix. Since A is a 4X5 matrix of rank 2, its nullity is 5 -2 = 3. Thus, the equation AX = 0 has 3 linearly independent solutions.

Let P be the matrix with columns X₁ = (17, 6,-13, 1,2)^T , X₂ = ( 1.2,3,4,5)^T and X₃ = ( 5,4,3,2,1)^T. Then the RREF of P ( on using a row-echelon reduction calculator) is R which has the columns Y₁ = ( 1, 0,0,0,0)^T , Y₂ = ( 0,1,0,0,0)^T and Y₃ = (0,0,1,0,0)^T . Apparently, Y₁ , Y ₂ , and Y₃ are linearly independent. Therefore, X ₁, X₂, and X₃ are also linearly independent. Thus, all other solutions of the equations AX = 0 are linear combinations of X ₁, X₂, and X₃ ( as the nullity of A is 3, there are only 3 linearly independent solutions of AX = 0). Thus, X ₁, X₂, and X₃   span the solution space to the system AX = 0.

(b) Let B be a 4 x 5 matrix of rank 3. Then, as per the rank-nullity theorem, the nullity of B is 5 -3 = 2. Thus, there can be only 2 linearly independent vectors in the solution space of the system BX = 0. Since X ₁, X₂, and X₃ are linearly independent, therefore.. all three of X ₁, X₂, and X₃ cannot satisfy the equation BX = 0. Thus, it is not possible to find a 4x5 matrix B of rank 3 such that BX_i = 0 will be satisfied for I = 1,2,3.