Dot Product Operator is Commutative

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Theorem

Let $\mathbf u, \mathbf v$ be vectors in a vector space $\mathbf V$ of $n$ dimensions.

Then:

$\mathbf u \cdot \mathbf v = \mathbf v \cdot \mathbf u$


Proof 1

\(\ds \mathbf u \cdot \mathbf v\) \(=\) \(\ds \sum_{i \mathop = 1}^n u_i v_i\) Definition of Dot Product
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n v_i u_i\) Real Multiplication is Commutative
\(\ds \) \(=\) \(\ds \mathbf v \cdot \mathbf u\) Definition of Dot Product

$\blacksquare$


Proof 2

The following proof is valid in the context of a Cartesian $3$-space:


\(\ds \mathbf u \cdot \mathbf v\) \(=\) \(\ds \norm {\mathbf u} \norm {\mathbf v} \cos \angle \mathbf u, \mathbf v\) Definition of Dot Product
\(\ds \) \(=\) \(\ds \norm {\mathbf v} \norm {\mathbf u} \cos \angle \mathbf u, \mathbf v\) Real Multiplication is Commutative
\(\ds \) \(=\) \(\ds \norm {\mathbf v} \norm {\mathbf u} \cos \angle \mathbf v, \mathbf u\) Cosine Function is Even
\(\ds \) \(=\) \(\ds \mathbf v \cdot \mathbf u\) Definition of Dot Product

$\blacksquare$


Examples

Example: $\paren {2 + 5 i} \circ \paren {3 - i} = \paren {3 - i} \circ \paren {2 + 5 i}$

Example: $\paren {2 + 5 i} \circ \paren {3 - i}$

Let:

$z_1 = 2 + 5 i$
$z_2 = 3 - i$

Then:

$z_1 \circ z_2 = 1$

where $\circ$ denotes (complex) dot product.


Example: $\paren {3 - i} \circ \paren {2 + 5 i}$

Let:

$z_1 = 3 - i$
$z_1 = 2 + 5 i$

Then:

$z_1 \circ z_2 = 1$

where $\circ$ denotes (complex) dot product.


As can be seen:

$\paren {2 + 5 i} \circ \paren {3 - i} = \paren {3 - i} \circ \paren {2 + 5 i}$

$\blacksquare$


Sources

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