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.

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