Question about transpose of quaternion matrix If $A$ is a matrix with entries in the quaternions and $q$ is a quaternion is
$(qA)^T = q A^T$ or $(qA)^T = \overline{q}A^T$?
 A: You can certainly define an operator $A^t_{ij} = A_{ji}$, but you probably want the conjugate transpose: $A^\dagger_{ij} = \bar{A}_{ji}$. The point is that $A^\dagger$ should be the transpose of $A$ with respect to the inner product on the space of quaternions, which is $\langle{x,y}\rangle = x\bar{y}$, not $xy$. (The latter is not positive-definite.) The operator $A^\dagger$ satisfies $\langle Ax, y\rangle = \langle x, A^\dagger y\rangle$; the operator $A^t$ does not.
A: If you are using a normal transpose it is $(qA)^T=qA^T$. If you are using a conjugate transpose (=Hermitian transpose) then it is $(qA)^H=\overline{q}A^H$.
See here for conjugate transpose.

On your follow-up question in the comments: $(AB)^T=B^TA^T$ doesn't hold if the entries of your matrix don't live in a commutative ring. So no, it does not hold in the quaternion case.
For example $$\left(\begin{pmatrix}i&0\\0&0\end{pmatrix}\begin{pmatrix}j&k\\0&0\end{pmatrix}\right)^T=\begin{pmatrix}k&-j\\0&0\end{pmatrix}^T=\begin{pmatrix}k&0\\-j&0\end{pmatrix}$$ But $$\begin{pmatrix}j&k\\0&0\end{pmatrix}^T\begin{pmatrix}i&0\\0&0\end{pmatrix}^T=\begin{pmatrix}j&0\\k&0\end{pmatrix}\begin{pmatrix}i&0\\0&0\end{pmatrix}=\begin{pmatrix}-k&0\\j&0\end{pmatrix}$$
