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I'm reading a paper now that works with operators on the vector space of real $d\times d$ matrices (such an operator can be thought of as a 4-dimensional tensor). The definition of positive-definiteness in this setting would seem to be the analogue of the usual one: we have the standard inner product $\langle A,B\rangle = \text{tr}(A^TB)$ and we say the operator $\mathcal{M}$ is positive-definite if for all matrices $A$ we have $\text{tr}(A^T\mathcal{M}A)>0$.

However, the definition this paper uses is that the above holds for all symmetric matrices $A$. What's the difference (if any?)? Why should this be the right definition (and is it?)?

More generally, is there a good reference on linear algebra over such spaces with special structure as the space of matrices?

amakelov
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There is no difference between your definition and the definition given in the paper. The reason is the following.

Given any matrix $T$ you can write it as $T=PO$ where $P$ is a positive matrix and $O$ is an orthogonal matrix. This is known as polar decomposition. Usually the proof is given for complex matrices but the same proof works for real matrices as well. See this post for details.

Now in your situation if you start with any matrix $T$, you have $$tr(T^TMT)=tr([PO]^TM[PO])=tr\big(O^T\bigg(P^TMP\bigg)O\big)=tr(P^TMP).$$

The last equality is due to the fact that $tr(AB)=tr(BA)$ and $O^TO=OO^T=Id$ . Therefore it is enough to check the inequality for positive matrices.

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