Linear Algebra NullSpace, Rank

Question

Let A = (a1 a2 a3 a4 a5), where ai is the i-th column of A, for i = 1,..., 5. It is known that

a1 = (1 -1 0 0), a3 = (-2 0 1 1), a5 = (-3 3 4 3).

Suppose A is row equivalent to R = (1 2 -1 2 -1; 0 1 -1 1 -1; 0 0 0 1 1; 0 0 0 0 0).

(a) What is the rank of A? Explain how you derive your answer.

(b) Show that { (2 2 2 1), (11 1 2 0), (5 4 3 2) } is a basis for the column space of A.

(c) Which of the following vectors v1 = (1 4 -3 7 0), v2 = (1 6 -5 3 1), v3 = (2 3 -1 1 -3), v4 = (1 1 0 4 3).

belongs to the row space of A? Explain how you derive your answer.

(d) Find a basis for the nullspace of A. Which is the nullity of A?

Answer 1

(a) The rank of a matrix is equal to the number of non-zero rows in its row echelon form. Since the row echelon form ( R ) has three non-zero rows, the rank of ( A ) is 3.(b) To show that ( {(2, 2, 2, 1), (11, 1, 2, 0), (5, 4, 3, 2)} ) is a basis for the column space of ( A ), we need to show that these vectors are linearly independent and span the column space of ( A ).(c) To determine which vector belongs to the row space of ( A ), we can multiply each vector by ( A ) and see if the result is a linear combination of the rows of ( A ).(d) To find a basis for the nullspace of ( A ), we need to solve the homogeneous system ( Ax = 0 ). Then, the nullity of ( A ) is equal to the number of free variables in the solution to this system.