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Mar 3, 2021 at 21:53 comment added Will Sawin In concrete terms, take $ c = \lambda + i \mu$ and $d = \lambda- i \mu$.
Mar 1, 2021 at 13:12 comment added YCor View an $n$-dimensional algebra as an $n^3$-tuple of numbers, namely coefficients of the law in a basis [and isomorphism means equivalence up to a suitable rule basis change]. The space $M_m(C)$ has a $C$-basis consisting of Hermitian matrices, in which the structure constants are real (and even integral). This basis is also a basis of the real algebra of Hermitian matrices, and the coefficients are just the same. This is all what you need.
Mar 1, 2021 at 12:41 history edited pi_1 CC BY-SA 4.0
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Mar 1, 2021 at 12:33 comment added pi_1 @dodd: They have the same dimension over $\mathbb C$ ($Herm(m,\mathbb C)$ is complexified) $dim_{\mathbb C}(M_m(\mathbb C))=m^2$ and for $dim(Herm(m,\mathbb C)\otimes_{\mathbb R}\mathbb C$, there the $m$ complex numbers on the diagonal and the $\frac{m(m-1)}{2}$ pair of complex numbers outside the daigonal (see for example p.158 of Analysis on symmetric cones by Faraut and Koranyi)
Mar 1, 2021 at 12:28 comment added pi_1 @Ycor: Thank you for your answer but I cannot see it: (complexified) hermitian matrices are of the form (for example $2\times 2$ $\left(\begin{smallmatrix}\alpha &a+Ib \\a-Ib &\beta\end{smallmatrix}\right)$ where $I$ stands for $i\otimes_{\mathbb R} \mathbb C$ and $\alpha,\beta,a,b\in \mathbb C$ but the $2\times 2$-matrices are just of the form $\left(\begin{smallmatrix}m &n \\o &p\end{smallmatrix}\right)$. What I would like to see is an expression of $m,n,o,p$ in terms of $\alpha,\beta, a,b$.
Mar 1, 2021 at 6:33 review Close votes
Mar 18, 2021 at 3:03
Mar 1, 2021 at 6:15 comment added YCor Try the canonical map induced by inclusion.
Mar 1, 2021 at 1:57 history asked pi_1 CC BY-SA 4.0