📂Linear Algebra

Orthogonal Complement of a Subspace

Definition¹

In the set

$$W^{\perp} = \left\{ \mathbf{v} \in V \ : \left\langle \mathbf{v} , \mathbf{w} \right\rangle = 0,\quad \forall \mathbf{w} \in W \right\}$$

for the subspace $W$ of a vector space $V$, it is called the orthogonal complement of $W$. Here $\langle , \rangle$ is the inner product.

Explanation

In other words, $W^{\perp}$ is a collection of vectors that are orthogonal to every element of $W$. The symbol $^{\perp}$ is abbreviated as perp[펍] for perpendicular. Because it’s a literal definition, it’s easily understood, especially in Euclidean spaces. For a simple example, if $W = \text{span} \left\{ (1,0,0) , (0,1,0) \right\}$ is given by $\mathbb{R}^{3}$ then it becomes $W^{\perp} = \text{span} \left\{ (0,0,1) \right\}$.

Properties

Let $W$ and $R$ be subspaces of $\mathbb{R}^{n}$, denoted by $A \in M_{ m \times n}(\mathbb{R})$. Then, the following properties are satisfied:

[1] $( W^{\perp} ) ^{\perp} = W$

[2] $W \oplus W^{\perp} = \mathbb{R}^{n}$

[3] $R \subset W \iff W^{\perp} \subset R^{\perp}$

[4] $\mathcal{N} (A)^{\perp} = \mathcal{R} (A)$

[5] $\mathcal{C} (A)^{\perp} = \mathcal{N} (A^{T})$

[6] $\mathbb{R}^m = \mathcal{C} (A) \oplus \mathcal{N} (A^{T})$

[7] $\mathbb{R}^n = \mathcal{R} (A) \oplus \mathcal{N}(A)$

These properties can also be considered obvious when thinking about the definition of orthogonal complement. However, [2] can be a bit tricky and needs to be checked carefully. These properties, along with the properties of considering null space and column space together, are important.

While these properties may seem complex at first glance, fortunately, seeing $\perp$ as the reversed version of $T$ makes them not too hard to memorize. It’s like flipping $\mathcal{N}$ and $C$ inside and out without confusion.

On another note, the Rank-Nullity Theorem could also be considered as applying $\dim$ to [6], [7].

Theorem

If $W$ is a subspace of the inner space $V$, then the following holds:

(a) $W^{\perp}$ is a subspace of $V$.

(b) $W \cap W^{\perp} = \left\{ \mathbf{0} \right\}$

Proof

(a)

Since $\langle \mathbf{w}, \mathbf{0} \rangle = 0$ for all $\mathbf{w} \in W$, at least $\mathbf{0}$ is included in $W^{\perp}$. Since it is not an empty set, we just need to check if it is closed under addition and scalar multiplication. Let’s say $\mathbf{u}, \mathbf{v} \in W^{\perp}$ $k\in R$. Then, the following equation is satisfied:

$$\begin{align*} \langle \mathbf{u} + \mathbf{v}, \mathbf{w} \rangle =& \langle \mathbf{u}, \mathbf{w} \rangle + \langle \mathbf{v}, \mathbf{w} \rangle = 0 + 0 = 0 \\ \langle k \mathbf{u}, \mathbf{w} \rangle =& k \langle \mathbf{u}, \mathbf{v} \rangle = k \cdot 0 = 0 \end{align*}$$ Thus, $W^{\perp}$ is a subspace.

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(b)

Let’s say $\mathbf{v} \in W \cap W^{\perp}$. Then, $\mathbf{v}$ being orthogonal to $\mathbf{v}$ means

$$\langle \mathbf{v}, \mathbf{v} \rangle = 0$$

And the $\mathbf{v}$ satisfying this is only $\mathbf{0}$ by the definition of the inner product.

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