# Orthogonal Complements

Definition: Let $V$ be an inner product space., and let $U$ be a subset of vectors from $V$. The Orthogonal Complement of $U$ is the set of vectors $v \in V$ such that $v$ is orthogonal every vector $u \in U$, that is $U^{\perp} = \{ v \in V : \: <v, u> = 0, \forall u \in U \}$. |

*Take important note that $U$ need not be a subspace of $V$, but instead, only a subset of $V$ in order to define the orthogonal complement $U^{\perp}$.*

For example, consider the vector space $\mathbb{R}^2$ with the Euclidean inner product, and take $U$ to be the subset of $\mathbb{R}^2$ that form any line in $\mathbb{R}^2$. Any vectors that are parallel to this line are contained in $U$, and so $U^{\perp}$ would be the set of vectors in $\mathbb{R}^2$ that are perpendicular to the vectors parallel to this line.

We will now outline some very basic properties of the orthogonal complement of a subset in the following proposition.

Proposition 1: Let $V$ be an inner product space and let $U$ be a subset of vectors from $V$. Then:a) $U^{\perp}$ is a subspace of $V$.b) $\{ 0 \}^{\perp} = V$.c) $V^{\perp} = \{ 0 \}$.d) If $U_1$ and $U_2$ are subsets of $V$ such that $U_1 \subseteq U_2$ then $U_1^{\perp} \supseteq U_2^{\perp}$. |

**Proof of a)**Clearly $U^{\perp}$ is a subset of $V$. Let $v, w \in U^{\perp}$. Then $<v, u> = 0$ and $<w, u> = 0$ for all $u \in U$. Let $a, b \in \mathbb{F}$ and take the inner product of $av + bw$ with $u$ to get:

- Therefore $(av + bw) \in U^{\perp}$ and so $U^{\perp}$ is a subspace of $V$.

**Proof of b)**Note that $\{ 0 \}^{\perp}$ is the set of all vectors $v \in V$ such that $<v, 0> = 0$. But all vectors $v \in V$ have this property, so $\{ 0 \}^{\perp} = V$.

**Proof of c)**Note that $V^{\perp}$ is the set of all vectors $v \in V$ such that $<v, v> = 0$. But the only vector that has this property is the zero vector by the definiteness property of inner product spaces, so $V^{\perp} = \{ 0 \}$.

**Proof of d)**Let $U_1$ and $U_2$ be subsets of $V$ such that $U_1 \subseteq U_2$. Let $w \in U_2^{\perp}$. Then $w$ is a vector in $V$ such that $<w, u_2> = 0$ for all $u_2 \in U_2$. But all vectors in $u_1 \in U_1$ are contained in $U_2$. So $<w, u_1> = 0$ for all $u_1 \in U_1$. Thus $w \in U_1^{\perp}$ and so $U_1^{\perp} \supseteq U_2^{\perp}$. $\blacksquare$

Proposition 2: Let $V$ be an inner product space and let $U$ be a finite-dimensional subspace of $V$. Then $V$ is the direct sum of $U$ and $U^{\perp}$, that is $V = U \oplus U^{\perp}$. |

*Note that for Proposition 2 to hold, $U$ cannot only be a subset of $V$. We require that $U$ must be a subspace of $V$ otherwise the direct sum would make no sense. Furthermore, $U$ must be a finite-dimensional subspace, although, $V$ may be an infinite-dimensional subspace.*

**Proof:**Let $U$ be a subspace of $V$. To show that $V = U \oplus U^{\perp}$ we must show that $V = U + U^{\perp}$ and $U \cap U^{\perp} = \{ 0 \}$.

- Since $U$ is a finite-dimensional subspace, then we can let $\{ e_1, e_2, ..., e_n \}$ be an orthonormal basis of $U$. Let $v \in V$. Then $v$ can be written as:

- From above we see that $u \in U$ since $u$ is simply a combination of the orthonormal basis vectors of $U$. Furthermore, if we take the inner product of $w$ with $e_j$ for $j = 1, 2, ..., n$, then $<w, e_j> = <v, e_j> - <v, e_j>$ since $\{ e_1, e_2, ..., e_n \}$ are orthonormal vectors. Therefore the vector $w$ is orthogonal to $e_1, e_2, ..., e_n$, and so $w$ is orthogonal vector in the $\mathrm{span} (e_1, e_2, ..., e_n)$ i.e, each vector in $U$. Thus $w \in U^{\perp}$ and so $v = u + w$ where $u \in U$ and $w \in U^{\perp}$ so $V = U + U^{\perp}$.

- Now we will show that $U \cap U^{\perp} = \{ 0 \}$. Let $v \in U \cap U{\perp}$. Then $v \in U$ and $v \in U^{\perp}$ and so $v$ is a vector such that $<v, v> = 0$. But by the definiteness property of an inner product, we know that this happens if and only if $v = 0$ so $U \cap U^{\perp} \subseteq \{ 0 \}$. Furthermore, if $v \in \{ 0 \}$ then $v = 0$ and since $U$ and $U^{\perp}$ are both subspaces, we must have that $0 \in U$ and $0 \in U^{\perp}$. Therefore $0 \in U \cap U^{\perp}$ and $\{ 0 \} \subseteq U \cap U^{\perp}$. Thus $U \cap U^{\perp} = \{ 0 \}$. $\blacksquare$

Corollary 1: Let $V$ be an inner product space and let $U$ be a finite-dimensional subspace of $V$. Then $U = \left ( U^{\perp} \right )^{\perp}$. |

- To prove Corollary 1, all we need to do is show that $U \subseteq \left ( U^{\perp} \right )^{\perp}$ and $\left ( U^{\perp} \right )^{\perp} \subseteq U$.

- Let $u \in U$. Then we have that $<u, v> = 0$ for all vectors $v \in U^{\perp}$ and so since $u$ is orthogonal to every vector $v \in U^{\perp}$ then we have that $u \in \left ( U^{\perp} \right )^{\perp}$ and so $U \subseteq \left ( U^{\perp} \right )^{\perp}$.

- Now let $u \in \left ( U^{\perp} \right )^{\perp}$. From Proposition 2, we have that $V = U \oplus U^{\perp}$ and so $u$ can be written as $u = v + w$ where $v \in U$ and $w \in U^{\perp}$. But $u - v = w$, so $(u - v) \in U^{\perp}$.

- Now we already have that $u \in \left ( U^{\perp} \right )^{\perp}$ and $v \in \left ( U^{\perp} \right )^{\perp}$ then $(u - v) \in \left ( U^{\perp} \right )^{\perp}$. Therefore $(u - v) \in U^{\perp} \cap \left ( U^{\perp} \right )^{\perp} and so [[$ u - v$ is orthogonal to itself, so $u - v = 0$ since the zero vector is the only vector that is orthogonal to itself. Therefore $u = v$. But $u = v \in U$ and so $u \in U$. Therefore $\left ( U^{\perp} \right )^{\perp} \subseteq U$.

- Thus $U = \left ( U^{\perp} \right )^{\perp}$. $\blacksquare$