Dot Product Operator is Commutative
Jump to navigation
Jump to search
This page has been identified as a candidate for refactoring of medium complexity. In particular: This is more general than $\R^n$, although the latter is the only context Proof 2 is compatible with. Until this has been finished, please leave {{Refactor}} in the code.
New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code. |
Theorem
Let $\mathbf u, \mathbf v$ be vectors in the real Euclidean space $\R^n$.
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
\(\ds \mathbf u \cdot \mathbf v\) | \(=\) | \(\ds \norm {\mathbf u} \norm {\mathbf v} \cos \angle \mathbf u, \mathbf v\) | 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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 22$: Dot or Scalar Product: $22.8$
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $111 \ \text{(a)}$