Given a quaternion $q$ and a complex number $\lambda$, for scalar multiplication you can either apply $\lambda$ to $q$ from the left (i.e. $\lambda q$) or from the right (i.e. $q\lambda$). This means you can interpret $\mathbb{H}$ as a left complex vector space or as a right complex vector space. Now consider the left and right multiplication maps
$$ L_p(x)=px, \qquad R_p(x)=xp. \tag{1}$$
Then $L_p(x\lambda)=L_p(x)\lambda$ for all complex numbers, so $L_p$ is a linear transformation of $\mathbb{H}$ as a right complex vector space. And $R_p(\lambda x)=\lambda R_p(x)$, so $R_p$ is a linear transformation of $\mathbb{H}$ as a left complex vector space.
However, $L_p$ is not linear if we treat $\mathbb{H}$ as a left complex vector space, and $R_p$ is not linear if we treat $\mathbb{H}$ as a right vector space. This is because $\mathbb{H}$ is not commutative. (Exercise.)
Using $\{1,\mathbf{j}\}$ as a basis for $\mathbb{H}$ as a complex vector space, you would write an arbitrary quaternion as $z+w\mathbf{j}$ if you're thinking left vector space, and write $z+\mathbf{j}w$ if you're thinking right vector space. (Also notice $w$ comes before $z$ in the alphabet, so we're backwards alphabetically.) You're thinking left vector space but you're examining the right-linear transformation $L_p$. That's problematic.
If you look at $L_p$ and think right vector space, you should get
$$ \begin{bmatrix} z & -\overline{w} \\ w & \phantom{-}\overline{z} \end{bmatrix}. \tag{2} $$
(Exercise.)
Representing quaternions as matrices using $R_p$ is problematic since $R_p\circ R_q\ne R_{pq}$. Indeed, this would be a function $\mathbb{H}\to M_2(\mathbb{C})$ which is not a homomorphism but rather an anti-homomorphism, i.e. it satisfies $R_p\circ R_q=R_{qp}$ (the order of multiplication is reversed). In order to turn it into a homomorphism proper, one must either (pre)compose it with an anti-automorphism of $\mathbb{H}$ or (post)compose with an anti-automorphism of $M_2(\mathbb{C})$. Quaternion conjugation satisfies $\overline{pq}=\overline{q}\,\overline{p}$ and matrix transpose satisfies $(AB)^T=B^TA^T$, so we can use these as anti-automorphisms.
In the case of quaternion conjugation, we can take $p=z+w\mathbf{j}$, then its conjugate $\overline{p}=\overline{z}-w\mathbf{j}$, then the matrix of $R_{\overline{p}}$ you can calculate (exercise) to be
$$ \begin{bmatrix} \overline{z} & \overline{w} \\ -w & z \end{bmatrix}, \tag{3} $$
and if instead you just calculated $R_p$'s matrix and took the transpose you'd get $(2)$ again.
Complex conjugation (applied entry-wise to a matrix) is an automorphism of $M_2(\mathbb{C})$, i.e. it satisfies $\overline{AB}=\overline{A}\,\overline{B}$, so we may compose it with any of the representations above to get another valid representation.
Yes. One can show the following 2 things:
Any octonion $x$ not in $\mathbb R$ generates a subalgebra $A$ isomorphic to $\mathbb C$.
For any octonion $y$, the subalgebra $B$ generated by $x$ and $y$ is associative.
This should be covered in treatments of composition algebra, e.g. it is probably in Springer-Veldkamp.
Hence if we take $y$ outside of $A$ in 1, by Frobenius' classification of $\mathbb R$-division algebras, $B$ must be isomorphic to $\mathbb H$ (a priori it may not be obvious $B$ is division, but you can get this a posteriori from Frobenius' theorem as adjoining inverses to $B$, which necessarily lie in $\mathbb O$, still gives you an associative algebra), and thus the 4-dimensional space spanned by $1, x, y, xy$. Since no finite collection of $\mathbb R^4$ subspaces of $\mathbb R^8$ cover $\mathbb R^8$, you get infinitely many copies of $\mathbb H$.
To say that you get (at least) the cardinality of the continuum copies of $\mathbb H$, it suffices to show
- For any nonreal octonion $z$, there is a subalgebra $B \simeq \mathbb H$ as above not containing $z$.
To see this, take $x$ as above not in $\mathbb R z$. Then $x$ and $z$ generate a space $B' \simeq \mathbb H$. Simply take $y \not \in B'$. Then the algebra $B$ generated by $x$ and $y$ cannot contain $B'$ since $B \ne B'$ but $\dim B = \dim B'$.
Best Answer
A $\mathbb R$-linear function $\phi:\mathbb O\to A$ to any real associative algebra which is multiplicative and unitary (so that $\phi(xy)=\phi(x)\phi(y)$ for all $x$, $y\in \mathbb O$, and $\phi(1)=1$) has to vanish on the bilateral ideal $I$ generated by the elements of the form $$(x\cdot y)\cdot z-x\cdot(y\cdot z)$$ with $x$, $y$, $z\in\mathbb O$. Now, this ideal is not zero because $\mathbb O$ is not associative, and therefore $I=\mathbb O$, because $\mathbb O$ has no non-trivial bilateral ideals (it is a division thing).
It follows that the map $\phi$ is in fact zero.