## Linear Algebra and Its Applications, Exercise 3.1.18

Exercise 3.1.18. Suppose that $S = \{0\}$ is the subspace of $\mathbb{R}^4$ containing only the origin. What is the orthogonal complement of $S$ ( $S^\perp$)? What is $S^\perp$ if $S$ is the subspace of $\mathbb{R}^4$ spanned by the vector $(0, 0, 0, 1)$?

Answer: Every vector is orthogonal to the zero vector. (In other words, $v^T0 = 0$ for all $v$.) So the orthogonal complement of $S = \{0\}$ is the entire vector space, or in this case $S^\perp = \mathbb{R}^4$.

If $S$ is spanned by the vector $(0, 0, 0, 1)$ then all vectors in $S$ are of the form $(0, 0, 0, d)$. Any vector whose last entry is zero is orthogonal to vectors in $S$. In other words, for vectors of the form $(a, b, c, 0)$ the inner product with a vector in $S$ is $a \cdot 0 + b \cdot 0 + c \cdot 0 + 0 \cdot d = 0+0+0+0 = 0$

The space of vectors of the form $(a, b, c, 0)$ is spanned by the vectors $(1, 0, 0, 0)$, $(0, 1, 0, 0)$, and $(0, 0, 1, 0)$, which are linearly independent and form a basis for the subspace. Thus $S^\perp$ is the subspace of $\mathbb{R}^4$ with basis vectors $(1, 0, 0, 0)$, $(0, 1, 0, 0)$, and $(0, 0, 1, 0)$.

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 .

This entry was posted in linear algebra and tagged . Bookmark the permalink.