Exercise 2.3.22. Given a vector space of dimension 7 and a subspace of of dimension 4, state whether the following are true or false:

1) You can create a basis for by adding three vectors to any set of vectors that is a basis for .

2) You can create a basis for by removing three vectors from any set of vectors that is a basis for .

Answer: 1) True. Suppose , , , and are a basis for . Per 2L (page 86) any linearly independent set in can be extended to a basis for by adding more vectors if necessary. The four vectors through are already linearly independent (since they are a basis) and hence can be extended by adding additional vectors to form a basis for .

More specifically, we can find three vectors , , and such that a) the three vectors are not in (and hence are linearly independent of through ), and b) the three vectors are linearly independent of each other. The resulting seven vectors are linearly independent. Since the dimension of is 7 these seven linearly independent vectors must be a basis for . (See exercise 2.3.15.)

2) False. Consider the vectors through with having a one in the position and zeros elsewhere. These vectors are linearly independent and span and hence are a basis for it.

Now suppose is the subspace of all vectors of the form . The vectors through are not in the subspace and hence cannot be part of a basis for it. Thus it is not possible to remove three vectors from the basis set through and form a basis for .

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.