Dot Product of Orthonormal Basis Vectors

Theorem
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be an orthonormal basis of a vector space $V$.

Then:
 * $\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \delta_{i j}$

where:
 * $\mathbf e_i \cdot \mathbf e_j$ denotes the dot product of $\mathbf e_i$ and $\mathbf e_j$
 * $\delta_{i j}$ denotes the Kronecker delta.

Proof
By definition of orthonormal basis:


 * $(1): \quad \tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is an orthogonal basis of $V$
 * $(2): \quad \norm {\mathbf e_1} = \norm {\mathbf e_2} = \cdots = \norm {\mathbf e_1} = 1$

From $(1)$ we have by definition that $\mathbf e_i$ and $\mathbf e_j$ are perpendicular whenever $i \ne j$

Thus, from Dot Product of Perpendicular Vectors, for all $i$ and $j$, when $i \ne j$:
 * $\mathbf e_i \cdot \mathbf e_j = 0$

From Dot Product of Vector with Itself, for all $i$ we have:
 * $\mathbf e_i \cdot \mathbf e_i = \norm {\mathbf e_i}^2$

From $(2)$, we have that:
 * $\norm {\mathbf e_i} = 1$

and so:
 * $\mathbf e_i \cdot \mathbf e_i = 1$

Putting this together, we have:


 * $\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \begin {cases} 1 & : i = j \\ 0 & : i \ne j \end {cases}$

The result follows by definition of the Kronecker delta.