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.

$\blacksquare$


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