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let $\mathbb{H}$ be the quaternions and $Mat(n,\mathbb{H})$ be the vector space of $n\times n$ matrices over $\mathbb{H}$. Let $H(n,\mathbb{H}):=\{ A\in Mat(n,\mathbb{H}): \overline{A}^t=A \}$ be the subspace of all hermitian matrices over $\mathbb{H}$.

1)My first question: the hermitian matrices are no vector space over $\mathbb{H}$, since if A is a hermitian matrix, so $i\cdot A$ is not hermitian anymore. What does the author really mean by this.

(...) 2)Later on, the author says, that if $A,B\in H(n,\mathbb{H})\Rightarrow Tr(AB^*)$ is real. But this is also not clear for me.

We have $Tr(AB^*)=Tr(AB)$ but if we consider two matrices like:

\begin{pmatrix} 1 & i\\ -i & 1\end{pmatrix} and \begin{pmatrix} 1 & j\\ -j & 1\end{pmatrix}

Then we see, that the Trace is not real. What's wrong in my understanding?

I hope, you can help me.

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Do you mean $A^*=\overline{A}^t$? –  rschwieb Jan 17 '13 at 15:16
    
Yes, I do. Thanks –  Braten Jan 17 '13 at 19:09
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1 Answer

For (1): A Hermitian matrix $A$ here must have a real diagonal, and the diagonal entries of $iA$ are not real.

For (2), it looks like you have a counterexample for the statement as written. It's a little strange that the $^*$ would appear if the statement were about Hermitian matrices... why write it at all?

It's true though that $tr(AA^*)$ is real for any $A$

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for(2):The * is there, since the author has define a inner product on $Mat (n,\mathbb{H})$ by (A,B)=Re(Tr(AB*)). Then he argues that if A,B belong to $H(n,\mathbb{H})$, then the trace is real... for(1): My question here is, why the author talks about the vector space of all hermitian matrices over $mathbb{H}$? I mean, it is obviously not! (e.g. i$\cdot$A) –  Braten Jan 17 '13 at 19:12
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OK, thanks. It does look familiar :) Without your book in front of me I can only ask silly questions... –  rschwieb Jan 17 '13 at 19:18
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@Braten Maybe he means as a real vector space? Spanning with real scalars would not cause problems. –  rschwieb Jan 17 '13 at 19:19
    
This was also my suggestion. Thank you, this would clear question 1. But what about the trace problem? Maybe tr(AB) is defined differently? –  Braten Jan 17 '13 at 19:29
    
@Braten Your counterexample, which seems to have trace $2(1-k)$, seems to preclude that... I wonder about this: it's true that the product of Hermitian matrices is Hermitian iff they commute with each other. If that were the case for these quaternionic matrices, maybe something similar holds, so that the trace will be the trace of a "hermitian matrix", which will be real. –  rschwieb Jan 17 '13 at 19:33
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