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:

$\mathbf u \cdot \mathbf u = \left\Vert{\mathbf u}\right\Vert^2$

where $\left\Vert{\mathbf u}\right\Vert$ is the length of $\mathbf u$.


Proof 1

Let $\mathbf u = \left({ u_1, u_2, \ldots, u_n }\right)$.

Then:

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

$\blacksquare$


Proof 2

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

$\blacksquare$


Note

Because this theorem is used to prove the general ($n$-dimensional) case of Cosine Formula for Dot Product, this proof is circular the way we have defined the dot product.

However, some texts use the cosine formula as the definition of the dot product and derive the sum of products form as a consequence.

The two definitions are equivalent, so we have included this proof to show how the statement would be proved from the cosine definition.