Final answer:
To find the basis for the null space of A = [2 2 -1 -2], we solve Ax = 0 for vector x. Through assigning free variables and simplifying, we identify a set of independent vectors that form the basis of the null space, consisting of {[-1, 1, 0, 0], [1/2, 0, 1, 0],[1, 0, 0, 1]}.
Step-by-step explanation:
To find a basis for the null space of the matrix A = [2 2 -1 -2], we need to solve the homogeneous equation Ax = 0, where x is a vector in R4. This involves setting up the equation as a system of linear equations and reducing it to its row-echelon form or reduced row-echelon form. However, since matrix A is a 1x4 matrix, the null space will consist of all vectors orthogonal to A. We are looking for vectors x = [x1, x2, x3, x4] such that 2x1 + 2x2 - x3 - 2x4 = 0.
Any vector in the null space can be expressed in terms of free variables. Since there is only one equation and four variables, we can assign free variables to x2, x3, and x4. Let's say x2 = s, x3 = t, and x4 = u. The equation simplifies to 2x1 = -2s + t + 2u, thus x1 = -s + (1/2)t + u. Substituting back into the vector form, the null space vectors could look like [-s + (1/2)t + u, s, t, u]. These can be broken down into three independent vectors based on the free variable we choose: [-1, 1, 0, 0], [1/2, 0, 1, 0], [1, 0, 0, 1]. Hence, the basis for the null space is {[-1, 1, 0, 0], [1/2, 0, 1, 0], [1, 0, 0, 1]}.