Linear Algebra and Its Applications, Review Exercise 2.2

Review exercise 2.2. Find a basis for a two-dimensional subspace of \mathbb{R}^3 that does not contain (1, 0, 0), (0, 1, 0), or (0, 0, 1).

Answer: One approach is to come up with a linear system that has a two-dimensional nullspace that excludes the coordinate vectors. For the nullspace to be two-dimensional there must be two free variables and only one basic variable; we assume that x_2 and x_3 are free variables and x_1 is the basic variable.

One possible system is x_1 + x_2 + x_3 = 0. The coordinate vectors are not solutions to this equation and hence are not part of the nullspace.

If we set x_2 = 1 and x_3 = 0 then we have x_1 = -1. If we set x_2 = 0 and x_3 = 1 then we again have x_3 = -1. The two vectors (-1, 1, 0) and (-1, 0, 1) are thus basis vectors for the nullspace.

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.

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