Prove that dimkerA dimkerArref n kSolutionLet dim Ker A

Prove that dim(ker(A)) = dim(ker(A_rref)) = n - k

Solution

Let dim Ker A = k. Let {v_1,...,v_k} be a basis for Ker A. Extend to a basis {v_1,...,v_k, v_(k+1),...,v_n} for E. Let S = {A_rrefv_(k+1),..., Arrefv_n}. Given any x in kerA_rref, there is a y in E such that A_rrefy = x. Write y = ?(i = 1, n) c_i v_i for some scalars c_i. Since v_1,..., v_k are vectors in Ker A, A_rrefy = ?(i = k+1, n) c_i Lv_i. So x = ?(i = k+1, n) c_i A_rrefv_i belongs to Span S. Since x was an arbitrary element of kerA_rref, it follows that S spans kerA_rref.

S is also linearly independent. For if ?(i = k+1, n) c_i A_rrefv_i = 0 is a linear dependence relation of elements of S, then by linearity of A, A(?(i = k+1, n) c_i v_i) = 0. In other words, ?(i = k+1, n) c_i v_i is in Ker A. Since {v_1,...,v_k} is a basis for Ker A, we can write ?(i = k+1, n) c_i v_i = ?(i = 1, k) d_i v_i for some scalars d_i. Let e_i = -d_i for 1 ? i ? k and e_i = c_i for k+1 ? i ? n. Then ?(i = 1, n) e_i v_i = 0. Since {v_1,...,v_n} is linearly independent, e_i = 0 for all i. In particular, c_(k+1) =