## 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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