Question

Remember that: A subspace is never empty, and is either the just the zero vector. i.e. [0), or has an infinite number of vectors A basis for a subspace is a set of t vectors. where t is the dimension of the subspace (usually a small number.) These vectors span the subspace and are linearly independent. This means that 0 can never part of a basis. The basis of the subspace (0) is empty. i.e. . A subspace can be written as all linear combination of its basis vectors, though it is usually enough to just give the basis. 1. Consider a 12-by-9 matrix of rank 7. What are the dimensions of the four fundamental subspaces associated with this matrix? 2. Find a basis for each of the four fundamental subspaces associated with this matrix: 6 12 5 3. Find a basis for each of the four fundamental subspaces associated with this matrix: 1 -1 2 4 7 0 3 6-6 2 2 5 10

Please answer questions 2&3. Thank you!

Answer 1

The basis four fundamental spaces are 1. Column space 2. Null space_defined for Matrix 'A' 3. Row space 4. Left Null space _defined for matrix 'A^T' A = [3 6 6 12 2 5] Converting this into echelon form by row operations gives A = [3 6 6 12 2 5] R_2 rightarrow R_2 - 2R_1 gives A = [3 0 6 0 2 1] R_1 rightarrow R-1 - 2 R_2 gives A = [3 0 6 0 2 1] = > R_1/3 gives A = [7 0 1 0 0 1] Here pivoted column is only 1^st column. So column space 'A' = (3 6) Null space of 'A' = > x_1 = -x_2 0 x_1 = -1 x_3 (x-1 x_2 x_3) = lambda_1 (0 -1 0) + lambda (0 0 -1) So null space of A = (0 -1 0), (0 0 -1) \r\n Now A^T = [3 6 5 6 12 5]