Triangle Inequality/Vectors in Euclidean Space

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Theorem

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

Let $\left\Vert{\cdot}\right\Vert$ denote vector length.

Then:

$\left \Vert {\mathbf{x} + \mathbf{y} }\right \Vert \le \left \Vert {\mathbf{x}}\right \Vert + \left \Vert { \mathbf{y} }\right \Vert$

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

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


Proof

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

We have:

\(\displaystyle \Vert \mathbf{x} + \mathbf{y} \Vert ^2\) \(=\) \(\displaystyle \left({\mathbf{x} + \mathbf{y} }\right) \cdot \left({\mathbf{x} + \mathbf{y} }\right)\) $\quad$ Dot Product of Vector with Itself $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \mathbf{x} \cdot \mathbf{x} + \mathbf{x} \cdot \mathbf{y} + \mathbf{y} \cdot \mathbf{x} + \mathbf{y}\cdot \mathbf{y}\) $\quad$ Dot Product Distributes over Addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \mathbf{x} \cdot \mathbf{x} + 2 \left({ \mathbf{x} \cdot \mathbf{y} }\right) + \mathbf{y}\mathbf{y}\) $\quad$ Dot Product Operator is Commutative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \Vert \mathbf{x} \Vert^2 + 2 \left({ \mathbf{x} \cdot \mathbf{y} }\right) + \Vert \mathbf{y} \Vert ^2\) $\quad$ Dot Product of Vector with Itself $\quad$


From the Cauchy-Bunyakovsky-Schwarz Inequality:

\(\displaystyle \vert \mathbf{x} \cdot \mathbf{y} \vert\) \(\le\) \(\displaystyle \Vert \mathbf{x} \Vert \Vert \mathbf{y} \Vert\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \mathbf{x} \cdot \mathbf{y}\) \(\le\) \(\displaystyle \Vert \mathbf{x} \Vert \Vert \mathbf{y} \Vert\) $\quad$ Negative of Absolute Value $\quad$
\(\displaystyle \Vert \mathbf{x} \Vert^2 + 2 \left({ \mathbf{x} \cdot \mathbf{y} }\right) + \Vert \mathbf{y} \Vert ^2\) \(\le\) \(\displaystyle \Vert \mathbf{x} \Vert^2 + 2 \left({\Vert \mathbf{x} \Vert \Vert \mathbf{y} \Vert}\right) + \Vert \mathbf{y} \Vert ^2\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({\Vert \mathbf{x} \Vert + \Vert \mathbf{y} \Vert}\right)^2\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \Vert \mathbf{x} + \mathbf{y} \Vert ^2\) \(\le\) \(\displaystyle \left({\Vert \mathbf{x} \Vert + \Vert \mathbf{y} \Vert}\right)^2\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \Vert \mathbf{x} + \mathbf{y} \Vert\) \(\le\) \(\displaystyle \Vert \mathbf{x} \Vert + \Vert \mathbf{y} \Vert\) $\quad$ taking the square root of both sides $\quad$

$\blacksquare$


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

\(\displaystyle \left \Vert {\mathbf v + \mathbf w } \right \Vert\) \(=\) \(\displaystyle \left \Vert {\lambda \mathbf w + \mathbf w } \right \Vert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left \Vert {\left({\lambda + 1}\right) \mathbf w } \right \Vert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda + 1}\right) \left \Vert{\mathbf w}\right\Vert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lambda \left \Vert{\mathbf w}\right\Vert + 1 \left\Vert{\mathbf w}\right\Vert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left \Vert{\lambda \mathbf w}\right\Vert + \left\Vert{1\mathbf w}\right\Vert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\Vert{\mathbf v}\right\Vert + \left\Vert{\mathbf w}\right\Vert\) $\quad$ $\quad$

$\Box$

Necessary Condition


Sources