Let $V$ be vector space. A *scalar product* is a map $\langle\ ,\ \rangle: V\times V\longrightarrow\mathbb{R}$ such that

SP 1. $\langle v,w\rangle=\langle w,v\rangle$

SP 2. $\langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle$

SP 3. If $x$ is a number, then

$$\langle xu,v\rangle=x\langle u,v\rangle=\langle u,xv\rangle$$

Additionally, we also assume the condition

SP 4. $\langle v,v\rangle>0$ if $v\ne O$

A scalar product with SP 4 is said to be *positive definite*.

*Remark*. If $v=O$, then $\langle v,w\rangle=0$ for any vector $w$ in $V$. This follows immediately from SP 3.

*Example*. Let $V=\mathbb{R}^n$ and define

$$\langle X,Y\rangle=X\cdot Y.$$

Then $\langle\ ,\ \rangle$ is a positive definite scalar product.

*Example*. Let $V$ be the function space of all continuous real-valued function on $[-\pi,\pi]$. For $f,g\in V$, we define

$$\langle f,g\rangle=\int_{-\pi}^{\pi}f(t)g(t)dt.$$

Then $\langle\ ,\ \rangle$ is a positive definte scalar product.

Using a scalar product, we can introduce the notion of *orthogonality* of vectors. Two vectors $v,w$ are said to be *orthogonal* or *perpendicular* if $\langle v,w\rangle=0$.

Let $S\subset V$ be a subspace of $V$. Let $S^{\perp}=\{w\in V: \langle v,w\rangle=0\ \mbox{for all}\ v\in V\}$.Then $S^{\perp}$ is also a subspace of $V$. (Check for yourself.) It is called the *orthogonal space* of $S$.

Define the *length* or the *norm* of $v\in V$ by

$$||v||=\sqrt{\langle v,v\rangle}.$$

It follows from SP 3 that

$$||cv||=|c|||v||$$

for any number $c$.

For vectors in $\mathbb{R}^2$ or $\mathbb{R}^3$, the vector projection of a vector $v$ onto another vector $w$ is

$$||v||\cos\theta\frac{w}{||v||}=\langle v,w\rangle\frac{w}{||w||^2}=\frac{\langle v,w\rangle}{\langle w,w\rangle}w$$

as seen in the above Figure. The number $c=\frac{\langle v,w\rangle}{\langle w,w\rangle}$ is called the component of $v$ along $w$. Note that the vector projection of $v$ onto $w$

$$\frac{\langle v,w\rangle}{\langle w,w\rangle}w$$

can still be defined in any vector space with a scalar product.

*Proposition*. The vector $v-cw$ is perpendicular to $w$.

*Proof*. \begin{align*}

\langle v-cw,w\rangle&=\langle v,w\rangle-c\langle w,w\rangle\\

&=\langle v,w\rangle-\langle v,w\rangle\\

&=0.

\end{align*}

*Example*. Let $V=\mathbb{R}^n$. Then the component of $X=(x_1,\cdots,x_n)$ along $E_i$ is

$$X\cdot E_i=x_i.$$

*Example*. Let $V$ be the space of continuous functions on $[-\pi,\pi]$. Let $f(x)=\sin kx$, where $k$ is a positive integer. Then

$||f||=\sqrt{\pi}$. The component of $g(x)$ along $f(x)$ is

$$\frac{\langle g,f\rangle}{\langle f,f\rangle}=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\sin kxdx.$$

It is called the *Fourier coefficient* of $g$ along $f$.

The following two inequalities are well-known for vectors in $\mathbb{R}^n$. They still hold in any vector space with a positive definite scalar product.

*Theorem* [*Schwarz Inequality*] Let $V$ be a vector space with a positive definite scalar product. For any $v,w\in V$,

$$|\langle v,w\rangle|\leq ||v||||w||.$$

*Theorem* [*Triangle Inequality*] Let $V$ be a vector space with a positive definite scalar product. For any $v,w\in V$,

$$||v+w||\leq ||v||+||w||.$$