Question

Tutorial on the Quaternions of Hamilton

Answer 1

Thank you Michele, I will give it a read! :)

Answer 2

thanks!! :) @Astrophysics

Answer 3

is this some sort of new language??? lol

Answer 4

no, the quaternions were introduced by W. R. Hamilton in the far 1833 :) lol

Answer 5

lol looks very complicated

Answer 6

no, they are not complicated, please study my definitions first @oleg3321

Answer 7

Very easy to understand, thanks again!

Answer 8

:) @Astrophysics

Answer 9

Interesting article. Well explained.
Hamilton was Ireland's most famous mathematician . Unfortunately his life was cut short by alcoholism.

Answer 10

Thanks! @welshfella

Answer 11

I remember my calc 3 teacher mentioning those to look at if you got bored..

thanks @Michele_Laino

Answer 12

thanks! :) @DanJS

Answer 13

Thank you @Michele_Laino .
I will definitely give it a read

Answer 14

Thanks! :) @arindameducationusc

Answer 15

quaternions aren't a field, they are a division ring

Answer 16

Quaternions are a field @oldrin.bataku

Answer 17

also you should introduce those other basis matrices \(\mathbf i,\mathbf k\) more naturally -- something like this: consider for \(z=a+bi\) we have: $$\begin{align*}a+bi\mapsto \begin{pmatrix}a+bi&0\\0&a-bi\end{pmatrix}&=\begin{pmatrix}a&0\\0&a\end{pmatrix}+\begin{pmatrix}bi&0\\0&-bi\end{pmatrix}\\&=a\begin{pmatrix}1&0\\0&1\end{pmatrix}+b\begin{pmatrix}i&0\\0&-i\end{pmatrix}\\&=a\cdot\mathbf1+b\cdot\mathbf i\end{align*}$$
and then we have that \(\mathbf i\cdot\mathbf j=\mathbf k\)

Answer 18

https://en.wikipedia.org/wiki/Quaternion#Noncommutativity_of_multiplication

>The fact that quaternion multiplication is not commutative makes the quaternions an often-cited example of a strictly skew field.

Answer 19

I wrote that formula:
\[{\mathbf{ij}} = - {\mathbf{ji}} = {\mathbf{k}}\]

Answer 20

https://en.wikipedia.org/wiki/Division_ring

>In abstract algebra, a division ring, also called a skew field, is a ring in which division is possible.
>Division rings differ from fields only in that their multiplication is not required to be commutative. ... Historically, division rings were sometimes referred to as fields, while fields were called “commutative fields”.

Answer 21

yes, I know you did, but you introduced \(\mathbf k\) as a matrix first without any motivation. the reason for its existence is that we have four linearly independent matrices \(\mathbf {1,i,j,ij}\) since the multiplication is bilinear, and then we just call \(\mathbf {ij}=\mathbf k\)

Answer 22

I never worked with non-commutative algebra

Answer 23

I meant I never cited non-commutative algebra

Answer 24

? fields have commutative multiplication by definition, whereas the quaternions \(\mathbb H\) are a non-commutative algebra over \(\mathbb R\) and a division ring, not a field

Answer 25

when I wrote a matrix in M2(C) it is supposed that the multiplication is non-commutative since as you know, the multiplication betwen matrices is non-commutative

Answer 26

lol, yes, clearly, and these matrices do not commute, so they cannot form a field. the quaternions are not a field.

Answer 27

:) right

Answer 28

For real man, tuts that are wrong are even worse. This is all simple definition stuff you are getting wrong...Please remove.

Answer 29

More precisely, I have used the term "field", since it is commonly used in mathematics literature, a more appropriate term can be "sfield" which indicate a non-commutative field.
I have supposed that, being the non-commutative characteristic evident, there was no need to use the term "sfield"

Answer 30

field is not commonly used in mathematics literature for a skew field, this isn't the 1800s anymore

Answer 31

it's called a division ring, that is the appropriate, correct, and commonly used term for algebraic structures like this