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Currently i am studying quaternions.

I do understand that i, j and k are imaginary numbers. so $i^2 = j^2 =k^2 = -1$. But I could not understand this: $$\begin{matrix}ij=k,&ji=-k,\\jk=i,&kj=-i,\\ki=j,&ik=-j\end{matrix}$$

Why is this? There seems no explanation why that is true. I would like to understand why that is true instead of just assuming that that is true.

Could somebody provide me some help?

mau
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Jordi Ozir
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    It's important to realize that the quaternion units are not really the same thing as imaginary numbers. Sure, the set ${ a + bi ;|; a, b \in \mathbb{R}}$ is a copy of $\mathbb{C}$, as are the sets ${a + bj ;|; a, b \in \mathbb{R}}$ and ${a + bk ;|; a, b \in \mathbb{R}}$. But the interactions such as $ij = k$ are not part of complex number arithmetic. – Shaun Ault Feb 06 '13 at 16:00
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    The key is that this is just a definition of how you multiply the $i,j,k$. There are subtle reasons for making them non-commutative, but it is probably too early to cover that. – Thomas Andrews Feb 06 '13 at 16:02
  • You say that it is just a definition. Do you mean with a definition that i^2=-1 or that that the bigest positive number is the infinity symbol? If yes than it would make sense. – Jordi Ozir Feb 06 '13 at 16:07
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    @user61231, there is no biggest positive number –  Feb 06 '13 at 16:44

4 Answers4

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$i,j,k$ are not imaginary numbers. Imaginary numbers arise only when you are talking about the complex plane $\mathbb{C}$, which has a very simple one to one mapping with the 2-D plane $\mathbb{R}^2$. Quaternions arise when you are talking about three dimensions, i.e. looking for solutions to $x^2+1 = 0$ in 3-D.

If you want to get a physical picture, consider $i$ as rotating a vector or a line segment in 3-D by $90^{\circ}$ taking X-axis as the axis of rotation. Similarly, $j,k$ correspond to rotations about Y and Z axes respectively. This is similar to imaginary number $i$, which corresponds to a right angle rotation in the complex plane. Since in 3-D there are more than one independent axes of rotation possible, 3 to be precise, there are 3 quaternions.

Now, two $90^{\circ}$ rotations about X, Y or Z axis will take the vector $\mathbf{x}$ to its mirror image $\mathbf{(-x)}$. S0, $i^2=j^2=k^2 =-1$.

A right angle rotation about $X$ axis followed by an equal amount of rotation about Y corresponds to an overall effective rotation of $90^{\circ}$ about Z axis. So, $ij = k$. Similarly, you can physically verify the quaternion multiplication laws.

Do not try to think of their multiplication as arithmetic. They are compositions of rotation operations. If this looks too confusing, a little background in group theory will bring you enough mathematical maturity to be comfortable with these.

dexter04
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  • Oke thank you very much. – Jordi Ozir Feb 06 '13 at 16:04
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    In the last sentence, I think you meant to restrict to $i,j$ and $k$. It's fine to think of multiplication between $i,j$ and $k$ as cross multiplication, but it is not okay to think of all quaternion multiplication that way. – rschwieb Feb 06 '13 at 16:06
  • Sorry. Updated the answer. – dexter04 Feb 06 '13 at 16:17
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    I think the geometric algebra interpretation of complex numbers and quaternions is the best, since it reveals more directly the fact that the "imaginary numbers" can be seen as encodings of rotations/reflections. Here is a pretty straightforward explanation. – amr Feb 07 '13 at 02:58
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    about this "A right angle rotation about XX axis followed by an equal amount of rotation about Y corresponds to an overall effective rotation of 90∘90∘ about Z axis. So, ij=kij=k.". the usual three axis vector x, y, z rotated in z axis results in x =(0 1 0) y = (-1 0 0) and z =(0 0 1). whereas x, y, z rotated in x axis results in x = (1 0 0) y = (0 0 1) z=(0 -1 0) and then y axis rotation results in x = (0 0 -1) y = (1 0 0) z = (0 -1 0) which is different from the k rotation. am i missing something here? – bysreg Oct 16 '16 at 23:19
  • your interpretation is perfectly reasonable but as @bysreg has noted above, a CCW rotation around the $X$ axis (attributed to $i$) followed by a CCW rotation around the $Y$ axis (attributed to $j$) does not equal a CCW rotation around the $Z$ axis (attributed to $k$), but a CW rotation! This is paradoxical as it suggests that $ij=-k$ and that $ijk=1$. The definition of the Hamilton Product seems to be axiomatic and implicitly based on an analogy with the cross product of the vector component of a quaternion (as an ordered pair of a scalar and a vector), hence the confusion. – Father Geppeto Jan 07 '22 at 18:16
  • This isn’t correct. Two 90 degree rotations don’t make - negative map - they make a 180 degree rotation which negates one plane and keeps one axis constant. Quaternions do correspond to rotations but it’s not that simple. – Eric Jun 10 '23 at 17:37
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The defining properties relating $i, j $ and $k$ are $$ i^2 = -1 \\ j^2 = -1 \\ k^2 = -1 \\ ijk = -1. $$ From these you get for example that $$\begin{align} i(ijk) &= -i \Rightarrow \\ i^2jk &= -i \Rightarrow\\ jk &= i. \end{align} $$ Likewise you can derive the other identities.

Thomas
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  • Are there any resources that explains this? – Jordi Ozir Feb 06 '13 at 16:13
  • @user61231: See for example: http://mathworld.wolfram.com/Quaternion.html – Thomas Feb 06 '13 at 16:14
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    These properties/definitions produce interesting and useful results, so we /axiomatically/ make them true for this structure. There is no deeper reason. We just want to study something that behaves in such a manner. – user173897 Dec 27 '14 at 05:35
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    so, does that mean there's no proof for ijk = -1? it kinda makes sense for the other three axioms (i^2 = sqr(sqrt(-1)) = -1) but i don't get why ijk has to equal to -1. is there any intuition behind it? – bysreg Oct 16 '16 at 23:07
  • @bysreg: It all depends on definitions. You could use another equivalent definition and then prove that $ijk = -1$. There is probably some nice intuition behind that $ijk = -1$, but I am not aware of it. One thought is that the requirements that $i^2 = j^2 = k^2$ is an attempt to extend the complex numbers. Now this doesn't give you a (commutative) field, but only a non-commutative division algebra. – Thomas Oct 17 '16 at 17:00
  • Why are jk and kj etc negatives of each other? – Aaron Franke Sep 19 '19 at 04:05
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You can derive the properties of quaternions through clifford algebra and the geometric product of vectors.

The geometric product works like so:

$$e_a e_b = \begin{cases} 1, & a = b \\ -e_b e_a, & a \neq b \end{cases}$$

where $a, b$ can be $x$, $y$, or $z$ as usual. This captures both the work of the cross product and the dot product in one product of basis vectors.

You can then identify

$$\begin{align*} i &= -e_y e_z \\ j &= -e_z e_x \\ k &= -e_x e_y \end{align*}$$

And then the properties of quaternions naturally follow.

$$i^2 = (-e_y e_z)(-e_y e_z) = (e_y e_z)(e_y e_z) = -e_y (e_z e_z) e_y = -e_y e_y = -1$$

And similarly for $j^2$ and $k^2$, as well as the $ijk$ product:

$$ijk = (-e_y e_z)(-e_z e_x)(-e_x e_y) = - e_y e_z e_z e_x e_x e_y = -e_y (e_z e_z)(e_x e_x) e_y = -e_y e_y = -1$$

This allows you to interpret quaternions in a very geometric way: the $i,j,k$ do not represent vectors, but rather oriented planes. It's just that in 3d each plane has a unique normal vector, so we often abuse this duality.

Muphrid
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  • But why are you identifying i as - e_y e_z ? By Clifford algebra e_y e_z = e_y . e_z + i | e_y X e_z | = | e_y | | e_z | exp (i * pi/2) = i – Subhadeep Ray Jan 15 '24 at 19:44
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enter image description here

It occurred to Hamilton during a walk to scratch into a bridge the following:

$$i^2 = j^2 = k^2 = ijk = -1.$$

From this we deduce by basic algebra:

  • $ij = -ijkk = k$
  • $jk = -iijk = i$
  • $jki = -1$
  • $ki = -jjki = j$
  • $i^{-1} = -i^{-1}ijk = -jk = -i$
  • $j^{-1} = -j^{-1}jki = -ki = -j$
  • $k^{-1} = -ijkk^{-1} = -ij = -k$
  • $ji = j^{-1} i^{-1} = (ij)^{-1} = k^{-1} = -k$
  • $kj = -i$ similarly
  • $ik = -j$ similarly

So we can provide complete arithmetic operations for the set numbers of the form $a + i b + j c + k d$:

$(a + i b + j c + k d) + (\alpha + i \beta + j \gamma + k \delta) = (a + \alpha) + i (b + \beta) + j (c + \gamma) + k (d + \delta)$

$\begin{array}? (a + i b + j c + k d) \cdot (\alpha + i \beta + j \gamma + k \delta) &=& a (\alpha + i \beta + j \gamma + k \delta) + i b (\alpha + i \beta + j \gamma + k \delta) + j c (\alpha + i \beta + j \gamma + k \delta) + k d (\alpha + i \beta + j \gamma + k \delta) \\ &=& a \alpha + i a \beta + j a \gamma + k a \delta + i b \alpha + i i b \beta + i j b \gamma + i k b \delta + j c \alpha + j i c \beta + j j c \gamma + j k c \delta + k d \alpha + k i d \beta + k j d \gamma + k k \delta \\ &=& (a \alpha - b \beta - c \gamma - d \delta) + i (a \beta + b \alpha + c \delta - d \gamma) + j (a \gamma - b \delta + c \alpha + d \beta) + k (a \delta + b \gamma - c \beta + d \alpha) \end{array}$

$$\frac{1}{a + i b + j c + k d} = \frac{a - i b - j c - k d}{(a + i b + j c + k d)(a - i b - j c - k d)} = \frac{a - i b - j c - k d}{a^2 + b^2 + c^2 + d^2}$$

  • I had travelled under that very bridge by rail so many times. https://www.buildingsofireland.ie/buildings-search/building/50060126/broome-bridge-ballyboggan-south-dublin-city – Rita Geraghty Dec 22 '19 at 18:07