Exercise 3.1.12. For the matrix

find a basis for the nullspace and show that it is orthogonal to the row space. Take the vector and express it as the sum of a nullspace component and a row space component .

Answer: We use Gaussian elimination to solve the system . We start by subtracting 1 times row 1 from row 2:

The resulting echelon matrix has two pivots, in columns 1 and 2, so and are basic variables and is a free variable. Setting from the second row we have or . From the first row we then have or . The vector is therefore a solution to and a basis for the nullspace.

Since is in the nullspace so is its negative, , which is an alternative basis for the nullspace. Note that the inner product of with is

and that the inner product of with is

The basis of the nullspace is thus orthogonal to each of the basis vectors for the row space, and . The row space as a whole consists of all linear combinations of those basis vectors, where and are scalars.

We then have

so that the basis of the nullspace is orthogonal to all vectors in the row space.

Note that we have

Since the vector is in the nullspace and the vector is in the row space (being the second row of ) we have where and .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.