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Algebraic Properties of the 2-norm of a Vector

Let column vectors $\vec{x}, \vec{y} \in R^{n \times 1}$ and scalar $a \in R$ . Then, the following are algebraic properties of 5.3 The 2-norm of a Vector:

Positivity: $| | \vec{x} | | \geq 0$ with $| | \vec{x} | | = 0$ iff $\vec{x} = \vec{0}$ .
Homogeneity: $| | a \vec{x} | | = | a | \cdot | | \vec{x} | |$ .
Triangle Inequality: $| | \vec{x} + \vec{y} | | \leq | | \vec{x} | | + | | \vec{y} | |$ .

Positivity

\begin{proof}
Consider $\vec{x} \in R^{n \times 1}$ and $y \in R$ .

| | \vec{x} | | = \sqrt{\vec{x} \cdot \vec{x}}

According to positivty of inner product,

\vec{x} \cdot \vec{x} \geq 0.

For all $y \in R$ ,

\sqrt{y} \geq 0.

Therefore,

\begin{aligned} ⟹ \sqrt{\vec{x} \cdot \vec{x}} \geq 0 \\ ⟹ | | \vec{x} | | \geq 0 \end{aligned}

\end{proof}

Homogeneity

\begin{proof}
Consider

a \vec{x} = a [\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}] = [\begin{matrix} a x_{1} \\ a x_{2} \\ ⋮ \\ a x_{n} \end{matrix}]

Taking the 2-norm of this, we get:

\begin{aligned} | | a \vec{x} | | & = \sqrt{(a x_{1})^{2} + (a x_{2})^{2} + \dots + (a x_{n})^{2}} \\ = \sqrt{a^{2} (x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2})} \\ = \sqrt{a^{2}} \cdot \sqrt{x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2}} \\ = | a | \cdot | | \vec{x} | | \end{aligned}

\end{proof}

Triangle Inequality

\begin{proof}
Recall that the square root function $f (t) = \sqrt{t}$ is increasing. Therefore, if $t_{1} \leq t_{2}$ , then $\sqrt{t_{1}} \leq \sqrt{t_{2}}$ . If we can prove $\sqrt{t_{1}} = | | \vec{x} + \vec{y} | | \leq | | x | | + | | y | | = \sqrt{t_{2}}$ , then we can conclude the Triangle Equality is true. Consider:

\begin{aligned} | | \vec{x} + \vec{y} | |^{2} & = (\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y}) & 2-norm squared = inner product \\ = \vec{x} \cdot (\vec{x} + \vec{y}) + \vec{y} \cdot (\vec{x} + \vec{y}) & Bilinearity of dot product \\ = \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{y} + \vec{y} \cdot \vec{x} + \vec{y} \cdot \vec{y} & Simplify \\ = | | \vec{x} | |^{2} + 2 \vec{x} \cdot \vec{y} + | | y | |^{2} & Rewrite inner product as sum \\ = | | x | |^{2} + \sum_{j = 1}^{n} 2 x_{j} y_{j} + | | y | |^{2} \\ \leq | | x | |^{2} + 2 | | x | | | | y | | \\ = (| | \vec{x} | | + | | \vec{y} | |)^{2} \end{aligned}

The second to last expression requires that we know

\sum_{j = 1}^{n} x_{j} y_{j} \leq | | \vec{x} | | | | \vec{y} | | .

This is known as the 5.8 Cauchy-Schwarz Inequality.
\end{proof}