The Euclidean Norm

# The Euclidean Norm

Recall from The Euclidean Inner Product page that if $\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$, then the Euclidean inner product $\mathbf{x} \cdot \mathbf{y}$ is defined to be the sum of component-wise multiplication:

(1)
\begin{align} \quad \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 + ... + x_ny_n = \sum_{i=1}^{n} x_iy_i \end{align}

We will now look at a very important operation related to the Euclidean inner product known as the Euclidean norm which we define below.

 Definition: If $\mathbf{x} = (x_1, x_2, ..., x_n) \in \mathbb{R}^n$ then the Euclidean Norm of $\mathbf{x}$ denoted $\| \mathbf{x} \|$ is defined to be $\| \mathbf{x} \| = \sqrt{ \mathbf{x} \cdot \mathbf{x}} = \left ( \sum_{i=1}^{n} x_i^2 \right )^{1/2}$.

The norm of $\mathbf{x}$ is therefore the square root of the Euclidean inner product of $\mathbf{x}$ with itself.

Note that when $n = 1$, the Euclidean norm of $x \in \mathbb{R}$ is $\| x \| = \sqrt{x \cdot x} = \sqrt{x^2} = \mid x \mid$, so the Euclidean norm of a real number is simply its absolute value. Let's instead look at a case where $n \geq 2$.

For example, consider the point $\mathbf{x} = (1, 2, 3, 4) \in \mathbb{R}^4$. Then we have that the norm of $\mathbf{x}$ is:

(2)
\begin{align} \quad \| \mathbf{x} \| = \sqrt{\mathbf{x} \cdot \mathbf{x}} = \sqrt{1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 4 \cdot 4} = \sqrt{1 + 4 + 9 + 16} = \sqrt{30} \end{align}

Let's now look at some important properties regarding the Euclidean norm of a point $\mathbf{x}$.

 Theorem 1: If $\mathbf{x} = (x_1, x_2, ..., x_n) \in \mathbb{R}^n$ and $a \in \mathbb{R}$ then: a) $\| \mathbf{x} \| \geq 0$. b) $\| a \mathbf{x} \| = \mid a \mid \| \mathbf{x} \|$. c) $\| \mathbf{x} \|^2 = \mathbf{x} \cdot \mathbf{x}$.
• Proof of a) Let $\mathbf{x} \in \mathbb{R}^n$.
• We note that $\mathbf{x} \cdot \mathbf{x} = \sum_{i=1}^{n} x_ix_i = \sum_{i=1}^{n} x_i^2$. Since $x_i^2 \geq 0$ for all $i \in \{1, 2, ..., n \}$ we have that $\mathbf{x} \cdot \mathbf{x} \geq 0$. So:
(3)
\begin{align} \quad \| \mathbf{x} \| = \sqrt{\mathbf{x} \cdot \mathbf{x}} \geq 0 \end{align}
• Proof of b) Let $\mathbf{x} \in \mathbb{R}^n$ and let $a \in \mathbb{R}$. Then:
(4)
\begin{align} \quad \| a \mathbf{x} \| = \sqrt{ (a \mathbf{x} \cdot a \mathbf{x} } = \sqrt{a^2 (\mathbf{x} \cdot \mathbf{x})} = \mid a \mid \sqrt{\mathbf{x} \cdot \mathbf{x}} = \mid a \mid \| \mathbf{x} \| \quad \blacksquare \end{align}
• Proof of c) Let $\mathbf{x} \in \mathbb{R}^n$. Then:
(5)
\begin{align} \quad \| \mathbf{x} \|^2 = \left ( \sqrt{\mathbf{x} \cdot \mathbf{x}} \right )^2 = \mid \mathbf{x} \cdot \mathbf{x} \mid \end{align}
• Since $\mathbf{x} \cdot \mathbf{x} \geq 0$ we have that:
(6)
\begin{align} \quad = \mid \mathbf{x} \cdot \mathbf{x} \mid = \mathbf{x} \cdot \mathbf{x} \quad \blacksquare \end{align}