Square of Vector Quantity in Coordinate Form
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Theorem
Let $\mathbf a$ be a vector in a vector space $\mathbf V$ of $n$ dimensions:
$\ds \mathbf a = \sum_{k \mathop = 1}^n a_k \mathbf e_k$
where $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.
Then:
- $\ds \mathbf a^2 = \sum_{k \mathop = 1}^n a_k^2$
where $\mathbf a^2$ denotes the square of $\mathbf a$.
Proof
By definition of square of $\mathbf a$:
- $\mathbf a^2 = \mathbf a \cdot \mathbf a$
By definition of dot product:
- $\ds \mathbf a \cdot \mathbf a = a_1 a_1 + a_2 a_2 + \cdots + a_n a_n = \sum_{k \mathop = 1}^n a_k^2$
$\blacksquare$
Sources
- 1927: C.E. Weatherburn: Differential Geometry of Three Dimensions: Volume $\text { I }$ ... (previous) ... (next): Introduction: Vector Notation and Formulae: Products of Vectors