Linear Algebra and Its Applications, Exercise 3.1.20

Exercise 3.1.20. Suppose S is a subspace of \mathbb{R}^n. Show that (S^\perp)^\perp = S. What does this mean?

Answer: We first consider the case where S = \{0\}; in other words, S contains only the zero vector. From exercise 3.1.18 we know that S^\perp = \{0\}^\perp = \mathbb{R}^n. The only vector that is orthogonal to all vectors in \mathbb{R}^n is the zero vector, so the zero vector is the only member of (S^\perp)^\perp = (\mathbb{R}^n)^\perp. We thus have S = (S^\perp)^\perp when S = \{0\}.

We next consider the case where S \ne \{0\}; in other words, S contains at least one nonzero vector (and thus has dimension of at least 1). Then there must exist a linearly independent set of one or more basis vectors v_1 through v_m that span S. Consider the m by n matrix A that has the basis vectors v_1 through v_m as its rows. The row space of A is then the set of all linear combinations of the basis vectors v_1 through v_m and is thus equal to the S itself.

Per 3D on page 138 (the Fundamental Theorem of Linear Algebra, part 2) the nullspace of A is the orthogonal complement of the row space and the row space of A is the orthogonal complement of the nullspace of A. But the row space is S, so the nullspace of A is the orthogonal complement S^\perp of S. The orthogonal complement of the nullspace of A is then (S^\perp)^\perp and is equal to the row space S.

So for all subspaces S in \mathbb{R}^n we have S = (S^\perp)^\perp.

To recap: For S = \{0\} we have S = (S^\perp)^\perp) trivially. Any nonzero subspace S is the row space of some matrix A and S^\perp is the nullspace of that matrix. The fact that S = (S^\perp)^\perp then follows from the fact that the row space of a matrix is the orthogonal complement of the nullspace and vice versa. The basis sets of S and S^\perp are mutually orthogonal, and together they form a basis set for \mathbb{R}^n. Thus no (nonzero) vector can be orthogonal to both S and S^\perp.

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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One Response to Linear Algebra and Its Applications, Exercise 3.1.20

  1. Math Tutor says:

    Thank you, thank you so much for sharing this exercise. It has really helped me and solved some of my confusion.

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