Why (pure) imaginary quaternions are the orthogonal complement to the real quaternions?

Question

I'm a layman of Lie Group and recently I'm trying to learn it with "Naive Lie Group" by John Stillwell (UTM series).

In the book, $\mathbf{i}, \mathbf{j}, \mathbf{k}, \mathbf{1}$ are defined as follows

$\mathbf{1} = \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix}$, $\mathbf{i} = \begin{bmatrix} 0 & -1\\1 & 0 \end{bmatrix}$, $\mathbf{j} = \begin{bmatrix} 0 & -i\\-i & 0 \end{bmatrix}$ and $\mathbf{k} = \begin{bmatrix} i & 0\\0 & -i \end{bmatrix}$ where $i$ is the square root of -1.

On the bottom part of page 12, it said that

$p = b\mathbf{i} + c\mathbf{j} + d\mathbf{k}$ where $b,c,d \in \mathbb{R}$ forms a three dimensional space that is the orthogonal complement to a$\mathbf{1}$, $a \in \mathbb{R}$.

However, IMHO, if this holds, then it means that for any $p = b\mathbf{i} + c\mathbf{j} + d\mathbf{k}$ and $q=a\mathbf{1}$, the product of these two matrices (p and q, in matrix form) are zero matrix.

If we let $a=b=c=d=1$, then

$p = \begin{bmatrix} i & -1-i\\1-i & -i \end{bmatrix}$ $q = \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} $ but $pq \neq \mathbf{0}$ ( $\mathbf{0}$ is a $2 \times 2$ zero matrix). So I don't know if there were any places I misunderstood. I checked some handouts/notes from Google, but as far as I've seen, none of them explain why pure imaginary quaternion space is orthogonal complement of real quaternion space.

I would highly appreciate it if anyone may give me some hints/explanations... Thank you very much for your reading.

Answer 1

In the context of Linear Algebra, we speak about two subspaces $V$ and $W$ of a given vector space being orthogonal only when we are dealing with an inner product. In the context of the quaternions, the natural inner product is\begin{align}\langle a+b\mathbf i+c\mathbf j+d\mathbf k,a'+b'\mathbf i+c'\mathbf j+d'\mathbf k\rangle&=\operatorname{Re}\bigl((a+b\mathbf i+c\mathbf j+d\mathbf k).(a'-b'\mathbf i-c'\mathbf j-d'\mathbf k)\bigr)\\&=aa'+bb'+cc'+dd'.\end{align}And, yes, with respect to this inner product, the spaces of real quaternions is orthogonal to the space of pure quaternions:$$\langle\lambda,a\mathbf i+b\mathbf j+c\mathbf k\rangle=\lambda\times0+0\times a+0\times b+0\times c=0.$$

Answer 2

In ring theory we may call elements of a ring orthogonal if they multiplie to zero. That is true. But in linear algebra we say two vectors are orthogonal if their inner product is zero. (Geometrically, this means the vectors are perpendicular if you were to draw them.) The space of $2\times 2$ complex matrices becomes a $8$-dimensional real inner product space with the inner product

$$ \langle \begin{bmatrix} a_1+a_2i & b_1+b_2i \\ c_1+c_2i & d_1+d_2i \end{bmatrix}, \begin{bmatrix} a_1'+a_2'i & b_1'+b_2'i \\ c_1'+c_2'i & d_1'+d_2'i \end{bmatrix} \rangle $$

$$ =a_1a_1'+a_2a_2'+b_1b_1'+b_2b_2'+c_1c_1'+c_2c_2'+d_1d_1'+d_2d_2'. $$

In other words, the real inner product that comes from the "most obvious" real vector space basis.

Personally I prefer to introduce quaternions as a sum of a scalar and a 3D vector, and then give the complex matrix representation later. In this context the inner product is just

$$ \langle w+xi+yj+zk,w'+x'i+y'j+z'k\rangle $$

$$ ww'+xx'+yy'+zz'. $$

Here it's rather obvious that scalars are orthogonal to vectors.