Triangle Inequality/Vectors in Euclidean Space

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Theorem

Let $\mathbf x, \mathbf y$ be vectors in the real Euclidean space $\R^n$.

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

Then:

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

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

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


Proof

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

We have:

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


From the Cauchy-Bunyakovsky-Schwarz Inequality:

\(\ds \size {\mathbf x \cdot \mathbf y}\) \(\le\) \(\ds \norm {\mathbf x} \norm {\mathbf y}\)
\(\ds \leadsto \ \ \) \(\ds \mathbf x \cdot \mathbf y\) \(\le\) \(\ds \norm {\mathbf x} \norm {\mathbf y}\) Negative of Absolute Value
\(\ds \leadsto \ \ \) \(\ds \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2\) \(\le\) \(\ds \norm {\mathbf x}^2 + 2 \paren {\norm {\mathbf x} \norm {\mathbf y} } + \norm {\mathbf y}^2\) multiply both sides with $2$, and add $\norm {\mathbf x}^2 + \norm {\mathbf y}^2$ to both sides
\(\ds \) \(=\) \(\ds \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2\)
\(\ds \leadsto \ \ \) \(\ds \norm {\mathbf x + \mathbf y}^2\) \(\le\) \(\ds \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2\)
\(\ds \leadsto \ \ \) \(\ds \norm {\mathbf x + \mathbf y}\) \(\le\) \(\ds \norm {\mathbf x} + \norm {\mathbf y}\) taking the square root of both sides

$\blacksquare$


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

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

$\Box$


Necessary Condition


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