Example Problems

Show that the inner product defined above for $\mathbb{R}^n$ as $\langle \vec{x}, \vec{y}\rangle = x_1y_1 + x_2y_2 + \cdots + x_ny_n$ fulfills the 4 properties of an inner product.

Let $\vec{x}, \vec{y}, \vec{z} \in \mathbb{R}^n$ and $\alpha \in \mathbb{R}$.

4. $\langle \vec{x}, \vec{x}\rangle = x_1^2 + x_2^2 + \cdots x_n^2$.

   Since every component is squared, we know that $\langle \vec{x}, \vec{x}\rangle \geq 0$. Let us check when $x_1^2 + x_2^2 + \cdots x_n^2 = 0$. Since we are adding each component squared, all components must be $0$ in order for this sum to be $0$ as well, i.e. $\vec{x}$ must be the zero vector. Therefore, $\langle \vec{x}, \vec{x}\rangle = 0 \iff \vec{x} = \vec{0}$.

Norm

The norm of a vector $\vec{x} \in V$ is its length. We denote the norm as $\|\vec{x}\|: V \rightarrow \mathbb{R}$.

For vectors in $\mathbb{R}^n$, to find the length of a vector $\vec{x} = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n\end{bmatrix}$, we use the Pythagorean theorem.

$\|\vec{x}\| = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} = \sqrt{\langle \vec{x}, \vec{x}\rangle}$

Therefore, we can express the inner product of $\vec{x}$ with itself as the square of its norm:

$\langle \vec{x}, \vec{x}\rangle = \|\vec{x}\|^2$

Properties

Norms have to fulfill 3 properties for $\vec{x}, \vec{y} \in V$ and $\alpha \in \mathbb{R}$:

1. Nonnegativity: $\|\vec{x}\| \geq 0$, where $\|\vec{x}\| = 0$ if and only if $\vec{x} = \vec{0}$

2. Scaling: $\|\alpha \vec{x}\| = |\alpha|\|\vec{x}\|$

3. Triangle inequality: $\|\vec{x} + \vec{y}\| \leq \|\vec{x}\| + \|\vec{y}\|$, where $\|\vec{x} + \vec{y}\| = \|\vec{x}\| + \|\vec{y}\|$ if and only if $\vec{x} = \alpha\vec{y}$ (i.e. when $\vec{x}$ and $\vec{y}$ are parallel)

Angle

We define the angle $\theta$ between 2 vectors $\vec{x}, \vec{y}$ using the following equation:

$\langle \vec{x}, \vec{y}\rangle = \|\vec{x}\|\|\vec{y}\|\cos\theta$

$\cos\theta = \frac{\langle \vec{x}, \vec{y}\rangle}{\|\vec{x}\|\|\vec{y}\|}$

Angle between 2 vectors