# Complex representation of a quaternionic matrix

It is evident that right module $\mathbb{H}^n$ is $\mathbb{C}$-linearly isomorphic to $\mathbb{C^{2n}}$ with corresponding isomorphism $\nu : \mathbb{C^{2n}} \to\mathbb{H}^n$ given by $\nu(a,b) = a + b\mathrm{j}$. This naturaly gives representation for any quaternionic matrix $M \in \mathcal{M}^{n \times m}(\mathbb{H})$ with two complex matrices $A,B \in \mathcal{M}^{n \times m}(\mathbb{C})$ as $M = A + B\mathrm{j}$.

It's assumed that complex matrix representing $\nu^{-1}M\nu$ in parallel with complex representation of quaternion numbers can be written in the form
$$\theta_{n,m}(M) = \theta_{n,m}(A+B\mathrm{j}) = \left[\begin{matrix} A & B \\ -\overline B & \overline{A} \end{matrix}\right]$$ where $\overline{A}$ is a complex conjugate. However, what i don't understand is there this conjugation came from and I need your help.

When I write $$\nu^{-1}M\nu(a,b) = \nu^{-1}(A +B\mathrm{j})(a + b\mathrm j) =\nu^{-1}\left(Aa + Ab\mathrm{j} + B\overline{a}\mathrm{j} -B\overline b\right) = \left( Aa - B\overline{b}, Ab + B\overline a \right)$$ I don't have any idea what to do with conjugates to show that this map even linear.

• Why do you need to know where the conjugation came from? Is it not enough to verify that the correspondence of multiplication and addition of two elements in $\mathbb{H}$ takes the underlying reals where they supposed to? – mathreadler Jul 5 '16 at 21:36
• At first i thought that where mast be simple connection between complex and quaternionic matrices as there are a simple connection between complex and quaternionic vector spaces and i expected to derive this matrix representation as solution. However, now it seems that conjecture is not true and that desired matrix representation can be achieved only with nonlinear change of basis, it is also possible that i maid a mistake and that actual result is trivial. I'm unsure. – Nik Pronko Jul 5 '16 at 22:17
• I think you are over-complicating and confusing things. Just build the matrix , assume $a_1+b_1i$ for $A_1$ and $c_1+d_1i$ for $B_1$ and the same for two other matrix $A_2, B_2$ and see what happens when you try matrix multiplication. Do they map as they are supposed to? For the change of basis, you may need to vectorize. It's not linear by matrices on the matrix per se, but if you start treating the matrix as a vector. – mathreadler Jul 5 '16 at 23:32
• Now I clearly can can see that this two maps are identical as the map which I will call $\theta_{n,m}$ is only $\mathbb{R}$-linear and not $\mathbb{C}$-linear. Real multiplication which you are suggesting clearly shows that $\theta_{n,m}(M) \neq \nu^{-1} M \nu$ with another representation through 4 real matrices. – Nik Pronko Jul 6 '16 at 19:18
• As Arctic Tern says in their deleted answer your mapping $\nu$ is not $\Bbb{C}$-linear because $\Bbb{C}$ is acting from the right. If you use $a+jb$ and $A+jB$ instead, then the conjugation will affect $A$ and $B$ as opposed to $a$ and $b$. The matrices must act on the vectors from the left so that they commute with the action of scalars of $\Bbb{H}$ from the right. – Jyrki Lahtonen Jul 7 '16 at 22:40

The key reason of my confusion was usage of left matrix on vector multiplication in a right module $\mathbb{H}^n$. As it turns out in right modules it would safe only to use corresponding matrix ring acting on the module from the right. Hence, if we assume that map $\nu$ also act from the right $(a,b)\nu = a + b\mathrm{j}$ we can get desired result $$(a,b)\nu M \nu^{-1} = (a +b\mathrm{j})(A +B\mathrm{j})\nu^{-1} = ( aA + aB\mathrm{j} +b\overline{A}\mathrm{j} -b\overline{B} )\nu^{-1} = (aA - b\overline{B},aB + b\overline{A})$$ which will yeld correct matrix representation if we assume that $aA \triangleq A^\top a^{(\top)}$.
Otherwise we can just define $\mathbb{H}^n$ as a left module and get similar result.
So we can actually think about $\theta_{n,m}$ as change of global charts if we think about quaternionc matrix M as a (nonlinear) function $\mathbb{H}^n \to \mathbb{H}^m$.