Standard Ordered Basis Vectors are Orthogonal
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Theorem
Let $\struct {\mathbf V, +, \circ}_{\mathbb F}$ be a vector space over a field $\mathbb F$, as defined by the vector space axioms.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis of $\mathbf V$.
Let $\mathbf e_i$ and $\mathbf e_j$ be elements of $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ such that $i \ne j$.
- $\mathbf e_i$ and $\mathbf e_j$ are orthogonal.
Proof
By definition of standard ordered basis:
- $\mathbf e_i$ is a vector whose $i$th component is $1_{\mathbb F}$ with all other components $0_{\mathbb F}$.
Hence:
- $\mathbf e_i \cdot \mathbf e_j = \delta_{i j}$
where $\delta_{i j}$ denotes the Kronecker delta.
That is:
- $i \ne j \implies \mathbf e_i \cdot \mathbf e_j = 0$
Hence the result by definition of orthogonal.
$\blacksquare$
Sources
- 1970: George Arfken: Mathematical Methods for Physicists (2nd ed.) ... (previous) ... (next): Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product