Statement

For F=RF=\mathbb R or C\mathbb C, equip D=F×FD=F\times F with conjugation (a,b)‾=(b,a)\overline{(a,b)}=(b,a). Projection onto the first matrix component defines a unital Jordan isomorphism

Herm⁡n(D) ≅ Mn(F)+,(X,XT)⟼X,\operatorname{Herm}_n(D)\ \cong\ M_n(F)^+, \qquad (X,X^T)\longmapsto X,

for every integer n≥1n\geq1. On the right, the product is X∘Y=(XY+YX)/2X\circ Y=(XY+YX)/2.

Verification

The equation (X,Y)∗=(X,Y)(X,Y)^*=(X,Y) is exactly Y=XTY=X^T. Thus the inverse map is X↦(X,XT)X\mapsto(X,X^T). Symmetrization of the product in the second component gives (XY+YX)T/2(XY+YX)^T/2, so projection preserves the Jordan product and the unit.

Catalogue instances

Over R\mathbb R, the coefficient algebra is . Over C\mathbb C, it is , whose composition conjugation exchanges the two factors. Consequently the complexification of the ordinary real Herm⁡n(C)\operatorname{Herm}_n(\mathbb C) is Mn(C)+M_n(\mathbb C)^+. It is not the ordinary conjugate-self-adjoint subspace over C\mathbb C.