Dot Product of Vector with Itself

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Theorem

Let $\mathbf u$ be a vector in the real vector space $\R^n$.

Then:

$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$

where $\norm {\mathbf u}$ is the length of $\mathbf u$.


Proof 1

Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_n}$.

Then:

\(\ds \mathbf u \cdot \mathbf u\) \(=\) \(\ds u_1 u_1 + u_2 u_2 + \cdots + u_n u_n\) Definition of Dot Product
\(\ds \) \(=\) \(\ds u_1^2 + u_2^2 + \cdots + u_n^2\)
\(\ds \) \(=\) \(\ds \paren {\sqrt {\sum_{i \mathop = 1}^n u_i^2} }^2\)
\(\ds \) \(=\) \(\ds \norm {\mathbf u}^2\) Definition of Vector Length in $\R^n$

$\blacksquare$


Proof 2

\(\ds \mathbf u \cdot \mathbf u\) \(=\) \(\ds \norm {\mathbf u} \norm {\mathbf u} \cos \angle \mathbf u, \mathbf u\) Cosine Formula for Dot Product
\(\ds \) \(=\) \(\ds \norm {\mathbf u}^2 \cos 0\) since the angle between a vector and itself is $0$
\(\ds \) \(=\) \(\ds \norm {\mathbf u}^2\) Cosine of Zero is One

$\blacksquare$


Also presented as

This can also be seen presented as:

$\norm {\mathbf u} = \paren {\mathbf u \cdot \mathbf u}^{1/2}$

or:

$\norm {\mathbf u} = u = \sqrt {\mathbf u \cdot \mathbf u}$

and so on.


Sources

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