Hamiltons quaternion rule states that ij=k and ji=-k. How can the commutative rule just be broken to make this true?

lilmoore11p8
2022-07-01
Answered

Hamiltons quaternion rule states that ij=k and ji=-k. How can the commutative rule just be broken to make this true?

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jugf5

Answered 2022-07-02
Author has **18** answers

"Commutative rule" (as you call it) cannot be "broken," since it is a property that some operations have and some don't. Nothing here is "broken." Multiplication of quaternions is just not commutative (or, in other words, does not satisfy the commutative property).

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Let $a=(a1,a2,a3)$ be a fixed vector in ${\mathbb{R}}^{3}$. Define the cross product $a\times v$ of $a$ and another vector $v=({v}_{1},{v}_{2},{v}_{3})\in {\mathbb{R}}^{3}$ as

$a\times v=det\left[\begin{array}{ccc}{e}_{1}& {e}_{2}& {e}_{3}\\ {a}_{1}& {a}_{2}& {a}_{3}\\ {v}_{1}& {v}_{2}& {v}_{3}\end{array}\right]$

Define a function $T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$ by $T(v)=a\times v$ for $v\in {\mathbb{R}}^{3}$.

a) Show that $T$ is a matrix transformation and calculate its representing matrix $M$

b) Find $\mathrm{ker}(T)$ and interpret its answer geometrically

c) find $\mathrm{range}(T)$ and interpret its answer geometrically

$a\times v=det\left[\begin{array}{ccc}{e}_{1}& {e}_{2}& {e}_{3}\\ {a}_{1}& {a}_{2}& {a}_{3}\\ {v}_{1}& {v}_{2}& {v}_{3}\end{array}\right]$

Define a function $T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$ by $T(v)=a\times v$ for $v\in {\mathbb{R}}^{3}$.

a) Show that $T$ is a matrix transformation and calculate its representing matrix $M$

b) Find $\mathrm{ker}(T)$ and interpret its answer geometrically

c) find $\mathrm{range}(T)$ and interpret its answer geometrically

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