The Euclidean Inner Product and Norm

# The Euclidean Inner Product

 Definition: Let $\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$. Then Euclidean Inner Product between $\mathbf{x}$ and $\mathbf{y}$ denoted $\mathbf{x} \cdot \mathbf{y}$ is defined to be $\displaystyle{\mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 + ... + x_ny_n = \sum_{i=1}^{n} x_iy_i}$.

Another term for the Euclidean inner product is simply "Dot Product".

Note that when $n = 1$ that the Euclidean inner product is simply the operation $\cdot$ of multiplication. Let's look at an example for when $n > 1$. Consider the points $\mathbf{x} = (1, 4, 9, 16), \mathbf{y} = (-4, -3, -2, -1) \in \mathbb{R}^4$. Then the dot product $\mathbf{x} \cdot \mathbf{y}$ is:

(1)
\begin{align} \quad \mathbf{x} \cdot \mathbf{y} = (1 \cdot (-4), 4 \cdot (-3), 9 \cdot (-2), 16 \cdot (-1)) = (-4, -12, -18, -16) \end{align}

We will now look at some nice properties of the Euclidean inner product that can be derived by the field axioms of $\mathbb{R}$.

 Theorem 1: If $\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n)$ and $a \in \mathbb{R}$ then: a) $\mathbf{x} \cdot \mathbf{y} = \mathbf{y} \cdot \mathbf{x}$. b) $(a \mathbf{x}) \cdot \mathbf{y} = \mathbf{x} \cdot (a \mathbf{y})$.
• Proof of a) Let $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$. Then:
(2)
\begin{align} \quad \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 + ... + x_ny_n \end{align}
• By the commutativity of multiplication, we have that:
(3)
\begin{align} \quad = y_1x_1 + y_2x_2 + ... + y_nx_n = \mathbf{y} \cdot \mathbf{x} \quad \blacksquare \end{align}
• Proof of b) Let $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ and $a \in \mathbb{R}$. Then:
(4)
\begin{align} \quad (a\mathbf{x}) \cdot \mathbf{y}) = (ax_1, ax_2, ... ax_n) \cdot (y_1, y_2, ..., y_n) = ax_1y_1 + ax_2y_2 + ... + ax_ny_n \end{align}
• Once again, by the commutativity of multiplication, we have that:
(5)
\begin{align} \quad = x_1(ay_1) + x_2(ay_2) + ... + x_n(ay_n) = \mathbf{x} \cdot (a \mathbf{y}) \quad \blacksquare \end{align}