Show that $\overline {x\cdot y} = \bar y \cdot \bar x,$ where $x,y \in \mathbb{H}.$

Question

In the book of Linear Algebra by Werner Greub, at page 209, it is given that

if $x,y \in \mathbb{H}$, then $$\overline {x\cdot y} = \bar y \cdot \bar x,$$ where $\cdot$ represents the product on quaternions and bar means the conjugate of the quaternion.

To prove this, I first tried to find the conjugate of the product $x \cdot y$; $$\overline{x\cdot y} = -(x,y)e + x\times y,$$ where $\times$ represents the cross product defined on $\mathbb{H}$. Then I have expanded the above expression with the decompositions $$x= \lambda_x e + x_0 \qquad y = \lambda_y e + y_0, \quad \text{where $x_0, y_0$ are in the complement space of $e$}.$$ Hence I get the expression

$$\overline{x\cdot y} = -e[\lambda_x \lambda_y + (x_0, y_0)] - e\times [\lambda_x y_0 - \lambda_y x] + \overline{x_0 \times y_0}.$$ However, I don't know how to find the conjugate of the term $x_0 \times y_0$ in the last expression. I mean if I can show that this cross product is in the orthogonal complement space of $e$, then I can say that $$\overline{x_0 \times y_0} = - x_0 \times y_0,$$ but I do not know whether this claim true or not, so I am basically stuck, and looking some help from you guys.

Answer 1

The easiest way i figured out to prove this fact is to carry out this computation for all the pairs of elements in $\{ 1, i, j, k \}$ which is trivial even if may be tedious

Once you have done it - assuming we are talking about real quaternion - the equality holds for all elements in $\mathbb{H}$ since

$$ \overline{(a_1v_1 + a_2v_2) \cdot (b_1w_1 + b_2w_2)} = \sum_{i=1}^2 \sum_{j=1}^2 a_i b_i \overline{ v_i \cdot w_j } $$

and

$$ \overline{(b_1w_1 + b_2w_2)} \cdot \overline{(a_1v_1 + a_2v_2)} = \sum_{i=1}^2 \sum_{j=1}^2 a_i b_j \overline{v_i} \cdot \overline{w_j} $$

For $a_i,b_j$ real numbers. The idea is to finally write $v$ and $w$ as the sum of their components.

Answer 2

You're treating every quaternion as a formal sum of a scalar and a vector. In this setting, the scalars are considered orthogonal to the vectors. The scalar component may be called the "real part" and the vector component may be called the "imaginary part."

Conjugation is defined as $\overline{r+\mathbf{v}}=r-\mathbf{v}$. So it's quite clear that $\overline{\mathbf{v}\times\mathbf{w}}=-\mathbf{v}\times\mathbf{w}$, because we have $\overline{\mathbf{u}}=-\mathbf{u}$ for any vector $\mathbf{u}$ and the cross product of two vectors is again a vector.

Answer 3

Although @anon gave a good intuitive answer, I still want to post my own solution to this problem because it took me a whole day to figure out what was the exact problem.

First of all, Mr.Greub defined the product $\cdot$ in $\mathbb{H}$, as a bilinear product on $\mathbb{H} \times \mathbb{H} \to \mathbb{H}$ as $$e\cdot x = x \cdot e = x \quad \forall x\in \mathbb{H},$$ and $$x_1\cdot y_1 = -(x_1, y_1)e + x_1\times y_1 \quad \forall x_1, y_1 \in \mathbb{H}_1,$$ where $\mathbb{H}_1$ denotes the orthogonal complement of $e$ in $\mathbb{H}$.

Now for a given $x \in \mathbb{H}$, we can decompose it as $$x= \lambda_x e + x_0 \quad x_0 \in \mathbb{H}_1$$, and since $\cdot$ is bilinear by its definition, $$x\cdot y = (\lambda_x e + x_0)\cdot (\lambda_y e + y_0) = \lambda_x \lambda_y \lambda_x y_0 + \lambda_y x_0 -(x_0, y_0)e + x_0 \times y_0, \quad \forall x, y \in \mathbb{H}.$$ And after this expansion, showing that $$\overline{x\cdot y} = \bar y \cdot \bar x$$ is pretty straightforward.