Triangle Inequality/Vectors in Euclidean Space

From ProofWiki
Jump to navigation Jump to search


Let $\mathbf x, \mathbf y$ be vectors in $\R^n$.

Let $\norm {\, \cdot \,}$ denote vector length.


$\norm {\mathbf x + \mathbf y} \le \norm {\mathbf x} + \norm {\mathbf y}$

If the two vectors are scalar multiples where said scalar is non-negative, an equality holds:

$\exists \lambda \in \R, \lambda \ge 0: \mathbf x = \lambda \mathbf y \iff \norm {\mathbf x + \mathbf y} = \norm {\mathbf x} + \norm {\mathbf y}$


Let $\mathbf x, \mathbf y \in \R^n$.

We have:

\(\displaystyle \norm {\mathbf x + \mathbf y}^2\) \(=\) \(\displaystyle \paren {\mathbf x + \mathbf y} \cdot \paren {\mathbf x + \mathbf y}\) Dot Product of Vector with Itself
\(\displaystyle \) \(=\) \(\displaystyle \mathbf x \cdot \mathbf x + \mathbf x \cdot \mathbf y + \mathbf y \cdot \mathbf x + \mathbf y \cdot \mathbf y\) Dot Product Distributes over Addition
\(\displaystyle \) \(=\) \(\displaystyle \mathbf x \cdot \mathbf x + 2 \paren {\mathbf x \cdot \mathbf y} + \mathbf y \cdot \mathbf y\) Dot Product Operator is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2\) Dot Product of Vector with Itself

From the Cauchy-Bunyakovsky-Schwarz Inequality:

\(\displaystyle \size {\mathbf x \cdot \mathbf y}\) \(\le\) \(\displaystyle \norm {\mathbf x} \norm {\mathbf y}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \mathbf x \cdot \mathbf y\) \(\le\) \(\displaystyle \norm {\mathbf x} \norm {\mathbf y}\) Negative of Absolute Value
\(\displaystyle \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2\) \(\le\) \(\displaystyle \norm {\mathbf x}^2 + 2 \paren {\norm {\mathbf x} \norm {\mathbf y} } + \norm {\mathbf y}^2\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \norm {\mathbf x + \mathbf y}^2\) \(\le\) \(\displaystyle \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \norm {\mathbf x + \mathbf y}\) \(\le\) \(\displaystyle \norm {\mathbf x} + \norm {\mathbf y}\) taking the square root of both sides


To prove that the equality holds if the vectors are scalar multiples of each other, assume:

$\exists \lambda \in \R, \lambda \ge 0: \mathbf v = \lambda \mathbf w$

Sufficient Condition

\(\displaystyle \norm {\mathbf v + \mathbf w}\) \(=\) \(\displaystyle \norm {\lambda \mathbf w + \mathbf w}\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {\paren {\lambda + 1} \mathbf w}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\lambda + 1} \norm {\mathbf w}\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \norm {\mathbf w} + 1 \norm {\mathbf w}\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {\lambda \mathbf w} + \norm {1 \mathbf w}\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {\mathbf v} + \norm {\mathbf w}\)


Necessary Condition