Orthogonal complement of a subspace

Let S be a subspace of \mathbb{R}^n. The orthogonal complement of S, denoted S^\perp, is the subspace of \mathbb{R}^n that contains the vectors orthogonal to all the vectors in S. If the subspace is described as the range of a matrix:

    \[ S = \{ Ax \: : \: x \in \mathbb{R}^n \}, \]

then the orthogonal complement is the set of vectors orthogonal to the rows of A, which is the nullspace of A^T.

Example: Consider the line in \mathbb{R}^3 passing through the origin and generated by the vector u = (1,2,3). This is a subspace of dimension 1:

    \[ S = \left\{ tu \: : \: t \in \mathbb{R} \right\} = \left\{ \left( \begin{array}{c} t \\ 2t \\ 3t \end{array} \right) \: : \: t \in \mathbb{R} \right\}. \]

To find the orthogonal complement, we find the set of vectors that are orthogonal to any vector of the form tu, with arbitrary t \in \mathbb{R}. This is the same set as the set of vectors orthogonal to u itself. So, we solve for u^T x = 0 with x \in \mathbb{R}^3:

    \[ x_1 + 2 x_2 + 3x_3 = 0. \]

This is equivalent to x_1 = -2x_2-3x_3. This equation characterizes the elements of the orthogonal complement S^\perp, in the sense that any x \in S^\perp can be written as

    \[ x = \left( \begin{array}{c} -2\alpha - 3\beta \\ \alpha \\ \beta \end{array} \right) = \alpha u + \beta v, \]

for some scalars \alpha, \beta, where

    \[ u = \left( \begin{array}{c} -2 \\ 1 \\ 0 \end{array} \right), \quad v = \left( \begin{array}{c} -3 \\ 0 \\ 1 \end{array} \right). \]

The orthogonal complement is thus the span of the vectors u, v:

    \[ S^\perp = \textbf{span}(u,v). \]

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