I tried to show by example that matrix multiplication for quaternionic matrices is is not necessarily $\mathbb H$-linear. If $A \in M_n(\mathbb H)$ is a quaternionic matrix and $x$ is a vector in $\mathbb H^n$ then we define multiplication as

$$ L_A: \mathbb H^n \to \mathbb H^n, (x_1,\dots,x_n) \mapsto (Ax^T)^T$$

where $^T$ denotes the transpose. I assumed that the transpose of a quaternionic matrix is really the conjugate transpose.

Now. The problem is: I did the computation several times and each time I got a slightly different result. But then I got the same result several times and I am posting this one here. Could please somebody tell me if I finally got it right?

Countereample: I chose $x=(j,j, \dots,) , y=(k,k,\dots), \lambda = i, \mu =j$ and $A=i\cdot I$. Then I computed

$$ L_A(\lambda x + \mu y) = (j+1, j+1, \dots)$$


$$ \lambda L_A(x) + \mu L_A (y) = (-j,-j,-j,\dots) + (1,1,1,\dots) = (1-j,1-j,1-j,\dots)$$

hence $L_A$ is not $\mathbb H$-linear. Did I get the computations right?

  • $\begingroup$ If you interpret vectors in $\mathbb{H}^n$ as column vectors rather than as row vectors, there is no need to mess about with transposes. $\endgroup$ – mweiss Jul 6 '14 at 15:54
  • $\begingroup$ @mweiss Yes I know but the book I'm reading (Tapp) defines multiplication like this (and it does not define the transpose at all). $\endgroup$ – learner Jul 6 '14 at 15:55

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