[Linear Algebra] 5. (a) Construct a matrix B whose null space consists of all linear combinations of vectors (2, 0, 7, 1) and (3, 1, −4, 0). (b) Express matrix B as a sum of two rank one matrices.

Sure, let's solve this step by step.

(a) The null space of a matrix B consists of all vectors x such that Bx = 0. If the null space of B is the set of all linear combinations of the vectors (2, 0, 7, 1) and (3, 1, -4, 0), then we can form a matrix A whose rows are these vectors:

A = [(2, 0, 7, 1), (3, 1, -4, 0)]

We can then take the null space of A to be the row space of B. To find B, we need to find a basis for the null space of A. This can be done by performing Gaussian elimination on A to bring it to row echelon form, and then reading off the basis for the null space.

After performing Gaussian elimination, we get:

A = [(1, 0, 7/2, 1/2), (0, 1, -7, -3/2)]

The null space of A is then the set of all vectors (x1, x2, x3, x4) such that x1 + 7/2x3 + 1/2x4 = 0 and x2 - 7x3 - 3/2x4 = 0. Solving these equations for x1 and x2 gives:

x1 = -7/2x3 - 1/2x4 x2 = 7x3 + 3/2x4

So the null space of A is the set of all vectors of the form (-7/2x3 - 1/2x4, 7x3 + 3/2x4, x3, x4). This is the row space of B, so we can take B to be the matrix:

B = [(-7/2, 7, 1, 0), (-1/2, 3/2, 0, 1)]

(b) A rank one matrix is a matrix that can be written as the outer product of two vectors. Since B has rank 2, it can be written as the sum of two rank one matrices. We can take these to be the outer products of the rows of B with the standard basis vectors. This gives:

B = [(-7/2, 7, 1, 0), (-1/2, 3/2, 0, 1)]

= [(-7/2, 7, 1, 0)]' * [1, 0] + [(-1/2, 3/2, 0, 1)]' * [0, 1]

= [(-7/2, 7, 1, 0), (0, 0, 0, 0)]

• [(0, 0, 0, 0), (-1/2, 3/2, 0, 1)]

So B can be expressed as the sum of the rank one matrices [(-7/2, 7, 1, 0)]' * [1, 0] and [(-1/2, 3/2, 0, 1)]' * [0, 1].