Are j and k on different imaginary planes than i?

I'm trying to understand Quaternions. So I understand that a Quaternion is written like $xi+yj+zk+w$. I also understand that $i^2 = j^2 = k^2 = ijk = -1$, and how that can be used to derive equations such as $ij = k$ and $jk = i$.

One things that confuses me is that $i$ is not equal to $j$ which is not equal to $k$. I can say $i^2 = -1$ and $j^2 = -1$ but I can't say $ij = -1$.

Correct me if I'm misunderstanding something, but why do they seem to have the same product when squared yet all three must be multiplied to equal the product of any one of them squared? Are they supposed to be on different imaginary planes?

Solution 1:

That is correct, $i$, $j$ and $k$ are contained in three separate complex planes contained in the quaternion numbers.

Just as the complex numbers $\mathbb{C}$ can be thought of as a 2-dimensional vector space over $\mathbb{R}$ with basis $1,i$, the quaternions $\mathbb{H}$ are a 4-dimensional vector space over $\mathbb{R}$ with basis $1,i,j,k$.

In particular, the plane spanned by $1,i$, the plane spanned by $1,j$, and the plane spanned by $1,k$ are different subplanes, any two of which intersect each other in the real line spanned by $1$.

So you can indeed think of these three planes as three separate copies of the complex numbers embedded in the quaternion numbers.

Solution 2:

One way to view quaternions is using matrices: $$ 1=\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix} $$ $$ i=\begin{bmatrix} 0&1&0&0\\ -1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0 \end{bmatrix} $$ $$ j=\begin{bmatrix} 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0 \end{bmatrix} $$ $$ k=\begin{bmatrix} 0&0&0&1\\ 0&0&-1&0\\ 0&1&0&0\\ -1&0&0&0 \end{bmatrix} $$ We can view this as $4$ orthogonal vectors which span a $4$ dimensional subspace of $\mathbb{R}^{16}$.

So yes, these can be viewed as $4$ orthogonal basis vectors, $3$ of which are orthogonal to the reals, thus "imaginary".