Let us consider a matrix algebra $Mat_{n\times n}(K)$, where $K$ is a field, $char K \neq 2.$

It is well-known that the axiomatization of commutator operation $[A,B]=AB-BA$ on matrix algebra leads us to the theory of Lie algebras. Axiomatization of $A\circ B= \frac{1}{2}(AB+BA)$ leads us to Jordan algebras.

Let us consider an operation $A \Box B= \frac{1}{2}(AB+BA^T),$ arising, for example, in control. How we can describe a class of algebras arising from axiomatization of such an operation?

**UPDATE** As it was shown by Pasha Zusmanovich below considering only $\Box$ leads us to a trivial variety(of course we can try to proceed to a quasivariety..).

But, if we add a transposition to the signature situation becomes much more interesting. First of all we have $(A\Box B)^T=A\Box B^T$ and the left unit $I\Box A= A$ Really, if we consider $T$-invariant subalgebras of some matrix algebra with $\circ$, than we can note that such algebras could be decomposed (as vector spaces) to the direct sum of Jordan algebra and Lie algebra -- symmetric and antisymmetric part,respectively. Axiomatizing this decomposition we get... Commutativity for symmetric part: $$ (A+A^T)\Box (B+B^T)=(B+B^T)\Box (A+A^T), $$ Power-associativity for symmetric part: $$ (A+A^T)\Box ((A+A^T)\Box (A+A^T))= ((A+A^T)\Box (A+A^T))\Box (A+A^T) $$ Jordan identity for symmetric part: $$ ((A+A^T)\Box (B+B^T))\Box ((A+A^T)\Box (A+A^T))= (A+A^T)\Box ((B+B^T)\Box ((A+A^T)\Box (A+A^T))) $$ Anticommutativity for antisymmetric part: $$ A\Box A+A^T\Box A^T=A\Box A^T+A^T\Box A $$ Lie identity for antisymmetric part: $$ (A-A^T)\Box ((B-B^T)\Box (C-C^T))+(C-C^T)\Box ((A-A^T)\Box (B-B^T))+(B-B^T)\Box ((C-C^T)\Box (A-A^T))=0 $$

Commutativity and power-associavity for symmetric part could be seen as averaged commutativity and averaged associativity and(!) commutativity, respectively. $$ A\Box B + A\Box B^T + A^T\Box B + A^T\Box B^T= B\Box A + B^T\Box A + B\Box A^T +B^T\Box A^T $$ $$ \sum_{\sigma\in S_3}\sum_{(i,j,k)\in (\varnothing,T)^3}A_{\sigma(1)}^{i}\Box(A_{\sigma(2)}^j\Box A_{\sigma(3)}^k) =\sum_{\sigma\in S_3}\sum_{(i,j,k)\in (\varnothing,T)^3}(A_{\sigma(1)}^{i}\Box A_{\sigma(2)}^j)\Box A_{\sigma(3)}^k $$

Did anyone consider something close to varieties of algebras with identities of that type?

half-transpose$0.5[b,a^T]+0.5(ba^T+a^Tb)$. Then, $a\square b = 0.5(f + f^T)$. – Suvrit Jul 25 '11 at 17:59