Dot Product of Vector with Itself/Proof 1

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