Square of Vector Quantity in Coordinate Form

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:


 * $\ds \mathbf a \cdot \mathbf a = a_1 a_1 + a_2 a_2 + \cdots + a_n a_n = \sum_{k \mathop = 1}^n a_k^2$