Proof for $\mathbf{v}\cdot\mathbf{v} = \| \mathbf{v} \|^2$

Question

Anyone have a good link for proving $\mathbf{v}\cdot\mathbf{v} = \| \mathbf{v} \|^2$?

I tried ProofWiki but the site doesn't seem to be responding right now.

Answer 1

Suppose $\textbf{v} = \left [ \begin{array}{c} v_1 \\ \vdots \\ v_n \\ \end{array} \right ]$. Then

$$ \textbf{v} \cdot \textbf{v} \;\; =\;\; \left [ \begin{array}{ccc} v_1 & \ldots & v_n \\ \end{array} \right ] \left [ \begin{array}{c} v_1 \\ \vdots \\ v_n \\ \end{array} \right ] \;\; =\;\; v_1^2 + \ldots + v_n^2 \;\; =\;\; ||\textbf{v}||^2. $$

I'm assuming that you meant in a finite-dimensional vector space, correct?