Complex-Linear Real Matrices

Question

I ve been studying Tapps's Matrix Groups. In chapter 2 he defines a map $ρ_n$:$M_n$($\mathbb{C}$)$\rightarrow$$M_{2n}$($\mathbb{R}$) as follow: for n=1 $$ ρ_1((a+bi))=\left[\begin{array}{cc} a & b \\ -b & a \\ \end{array} \right], $$ and $ρ_n$ is defined by applying $ρ_1$ to each position of a nxn complex matrix. Matrices contained in the image $Im$$ρ_n$ are called complex-linear real matrices.

So far so good. Now consider the matrix $J_{2n}$ =$ρ_n$($iI$). There is a comment that the transformation defined by that matrix in $R^{2n}$, lets denoted it by $R_i$ mimicks scalar multiplication by i in $C^{n}$. But in what sense? From my point of view we can consider $R_i$ acting on a vector as scalar multiplying it by i. (Note that $J_{2n}$$J_{2n}$=-I). Any other insights are more than welcome!

The usual term, by the way, is "entry" rather than "entrance". It's surprisingly confusing when the wrong word is used in this context. — Ben Grossmann, Mar 17 '20 at 20:42
It's not a direct answer to your question, but you might find this post about the map $\rho_n$ to be helpful — Ben Grossmann, Mar 17 '20 at 20:48
@Omnomnomnom I totally agree with your first comment, it was a rookie mistake..Thanks for the answer, I ll give that post a look!! — mits314, Mar 17 '20 at 20:58

score 1 · Answer 1 · answered Mar 17 '20 at 20:59

Now consider the matrix $J_{2n}$ =$ρ_n(iI)$. There is a comment that the transformation defined by that matrix in $R^{2n}$, lets denoted it by $R_i$ mimicks scalar multiplication by i in $C^{n}$. But in what sense?

To answer this question directly: one answer is that $\rho_n:M_{n}(\Bbb C) \to M_{2n}(\Bbb R)$ is an isomorphism of rings (or of $\Bbb R$-algebras more specifically). It follows that for any matrix $A + iB \in M_n(\Bbb C)$ (with $A,B$ real), we have $$ J_n \cdot \rho_n(A + iB) = \rho_n(iI) \rho_n(A + iB) = \rho_n[(iI)(A + iB)] = \rho_n[i(A + iB)]. $$ The same idea applies for the reverse multiplication, so that $\rho_n(A + iB) J_n = \rho_n[(i(A + iB))]$. Informally, we could say the following: $\rho_n(A + iB)$ is the "mapped" version of $A + iB$. Multiplying $\rho_n(A + iB)$ by $J_n$ yields the mapped version of $i(A + iB)$.

I see, thanks for the rigorous answer! – mits314 Mar 17 '20 at 21:08 — mits314, Mar 17 '20 at 21:08

Complex-Linear Real Matrices

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