## 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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