Inner Product Space

이 포스트에서 $V$는 $F$-벡터공간으로 취급한다.

Inner Product

Definition 1. An inner product on $V$ is a function $\langle \cdot, \cdot \rangle: V \times V \longrightarrow F$, such that $\forall x, y, z \in V$ and $\forall c \in F$, the following hold:
(a) $\langle x + z, y \rangle = \langle x, y \rangle + \langle z, y \rangle$.
(b) $\langle cx, y \rangle = c \langle x, y \rangle$.
(c) $\overline{ \langle x, y \rangle} = \langle y, x \rangle$
(d) $\langle x, x \rangle > 0$ if $x \neq \mathbf{0}$.
Then $(V, \langle \cdot, \cdot \rangle)$ is called an inner product space.

    일반적으로 내적은 위와 같이 정의되며, 무수히 많은 내적이 존재한다. 익히 알고 있는 벡터의 내적은 수많은 내적 중 하나이며, 예시에 소개되어 있다.

Theorem 1

Theorem 1. Let $(V, \langle \cdot, \cdot \rangle)$ be an inner product space. Then $\forall x, y, z \in V$ and $c \in F$, the following statements are true.
(a) $\langle x, y + z \rangle = \langle x, y \rangle + \langle x, z \rangle$.
(b) $\langle x, cy \rangle = \overline{c} \langle x, y\rangle$.
(c) $\langle x, \mathbf{0}\rangle = \langle \mathbf{0}, x\rangle = 0$.
(d) $\langle x, x\rangle = 0 \iff x = \mathbf{0}$.
(e) If $\langle x, y\rangle = \langle x, z\rangle, \forall x \in V$, then $y = z$.

Proof. (a) $\langle x, y+z \rangle = \overline{\langle y+z, x\rangle} = \overline{\langle y, x\rangle + \langle z, x\rangle} = \overline{\langle y, x\rangle} + \overline{\langle z, x\rangle} = \langle x, y\rangle + \langle x, z\rangle$.
(b) $\langle x, cy \rangle = \overline{\langle cy, x\rangle} = \overline{c \langle y, x\rangle} = \overline{c} \overline{\langle y, x\rangle} = \overline{c} \langle x, y\rangle$.
(c) $\langle x, \mathbf{0}\rangle = \langle x, \mathbf{0} + \mathbf{0}\rangle = \langle x, \mathbf{0}\rangle + \langle x, \mathbf{0}\rangle$. Then $\langle x, \mathbf{0} \rangle = 0$. Similarly, $\langle \mathbf{0}, x\rangle = 0$.
(d) If $x \neq \mathbf{0}$, then $\langle x, x\rangle > 0$. But since $\langle x, x\rangle = 0, x = \mathbf{0}$. If $x = \mathbf{0}$, $\langle x, x\rangle = 0$ by (c).
(e) Take $x = y - z$. Then $\langle x, y \rangle = \langle x, z\rangle \Longrightarrow \langle y - z, y - z\rangle = 0 \iff y - z = \mathbf{0} \iff y = z$. $\blacksquare$

Remark

Remark. Let $(V, \langle \cdot, \cdot \rangle)$ be an inner product space. Let $x, y \in V$.
(a) Define $||x|| = \sqrt{\langle x, x\rangle}$. Then $||x||$ is the norm.
(b) Define $d(x, y) = ||x - y||$. Then $d$ is the metric.

    내적이 주어지면 항상 위와 같이 놈과 거리를 정의할 수 있다. 내적 공간에서 놈과 거리는 위의 정의로 사용한다. 즉 내적 공간이면 놈 공간이자 거리 공간이다. 이와 관련하여 아래와 같은 유용한 부등식과 항등식들이 성립한다. 

Theorem 2

Theorem 2. Let $(V, \langle \cdot, \cdot \rangle)$ be an inner product space. Then $\forall x, y \in V, \forall c \in F$, the following statements are true.
(a) $|\langle x, y\rangle| \leq ||x|| \, ||y||$. (Cauchy-Schwarz Inequality)
(b) $||x + y|| \leq ||x|| + ||y||$. (Triangle Inequality)
(c) $| \, ||x|| - ||y|| \, | \leq ||x - y||$. (Reverse Triangle Inequality)
(d) $||x + y||^2 + ||x - y||^2 = 2||x||^2 + 2||y||^2$. (Parallelogram Law)
(e) $||x + y||^2 = ||x||^2 + ||y||^2$ if $x$ and $y$ are orthogonal. (Pythagorean Theorem)

Proof. (a) If $y = \mathbf{0}$, then the result is immediate. Assume that $y \neq \mathbf{0}$. For any $c \in F$ we have $0 \leq ||x - cy||^2 = \langle x-cy, x-cy\rangle$ = $||x||^2 - \overline{c}\langle x, y\rangle - c\langle y, x\rangle + |c|^2 \, ||y||^2.$ In particular, if we set $c = \frac{\langle x, y\rangle}{||y||^2}$, then $0 \leq ||x||^2 - \frac{|\langle x, y\rangle|^2}{||y||^2}$.
(b) $||x+y||^2 = \langle x+y, x+y\rangle = ||x||^2 + 2\mathfrak{R}\langle x, y\rangle + ||y||^2$ $\leq ||x||^2 + 2 |\langle x, y\rangle| + ||y||^2$ $\leq ||x||^2 + 2||x|| \cdot ||y|| + ||y||^2 = (||x|| + ||y||)^2$ by (a).
(c) $||y + (-x)|| \leq ||y|| + ||-x|| = ||y|| - ||x|| \Longrightarrow ||x|| - ||y|| \leq -||x - y||$ $\Longrightarrow | \, ||x|| - ||y|| \, | \leq ||x - y||$ by (b).
(d) $||x + y||^2 + ||x-y||^2 = \langle x+y, x+y\rangle + \langle x-y, x-y\rangle$ = $2||x||^2 + 2||y||^2$.
(e) Since $x$ and $y$ are orthogonal, $\langle x, y\rangle = 0$. Then we have $||x + y||^2 = ||x||^2 + ||y||^2$. $\blacksquare$

Remark

Remark. If $V$ has an inner product $\langle x, y\rangle$ and $W \leq V$, then $W$ is also an inner product space when the same function $\langle x, y\rangle$ is restricted to $x, y \in W$.

Examples of Inner Product

(1) For $x = (a_1, ..., a_n), y = (b_1, ..., b_n) \in F^n$, define $$\langle x, y\rangle = \sum_{i=1}^n a_i \overline{b_i}.$$ It is easily verified that $\langle \cdot, \cdot \rangle$ is the inner product. This $\langle \cdot, \cdot \rangle$ is called the standard inner product of $F^n$. (When $F = \mathbb{R}$, this standard inner product is usually called the dot product and is denoted $x \cdot y$.)
(2) For $A, B \in M_{n \times n}(F)$, define $$\langle A, B \rangle = \text{tr}(B^*A),$$ where $B^*$ is the adjoint of $B$. Then $\langle \cdot, \cdot \rangle$ is the inner product on $M_{n \times n}(F)$. This $\langle \cdot, \cdot \rangle$ is called the Frobenius inner product.