## 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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### 1 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.