Dot Product of Vector with Itself/Proof 1
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Theorem
- $\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$
Proof
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 $\R^n$ |
$\blacksquare$
Sources
- 1992: Frederick W. Byron, Jr. and Robert W. Fuller: Mathematics of Classical and Quantum Physics ... (previous) ... (next): Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product