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Real matrices that lie in the image of the inclusion homomorphism $\rho_n: M_n(\mathbb C) \to M_{2n}(\mathbb R)$ are called complex linear real matrices. It is easy to see that a real matrix is complex linear if and only if it commutes with $I = \rho_n(iI)$.

In analogy to this I am now studying the quaternionic inclusion $M_n(\mathbb H) \to M_{4n}(\mathbb R)$ using the inclusion $\psi_n: M_n(\mathbb H) \to M_{2n}(\mathbb C)$. If $i,j,k$ denote the unit quaternions then I want to find matrices $I$ and $J$ such that a real matrix is quaternionic linear if and only if it commutes with $I$ and $J$.

In $1$ dimension I tried $I=\rho_{2n} \circ \psi_n (iI)$ and $J=\rho_{2n} \circ \psi_n (jI)$ but the problem then is that $I^2 \neq -1$ and $J^2 \neq -1$.

Why does $I=\rho_{2n} \circ \psi_n (iI), J=\rho_{2n} \circ \psi_n (jI)$ not work in the quaternionic case? Is there an insightful geometric (or other) explanation? For the inclusion of complex matrices into real matrices setting $J=\rho_{2n} (iI)$ worked.

Edit For a definition of $\rho_n$:

define $\rho_n : M^n(\mathbb C) \to M^{2n}(\mathbb R)$ as $A_{ij}\mapsto \begin{array}{cc} a_{ij} & b_{ij} \\ -b_{ij} & a_{ij} \end{array}$ if $A_{ij}=(a_{ij} + i b_{ij})$

and $\color{blue}{\psi_n}: M^n(\mathbb H) \to M^{2n}(\mathbb C)$ as $A_{ij}\mapsto \begin{array}{cc} a_{ij} & b_{ij} \\ -\overline{b_{ij}} & \overline{a_{ij}} \end{array}$ if $A_{ij}=(a_{ij} + b_{ij}j)$

Edit 2 (in response to the anwer)

Let $$I = \rho(\color{blue}{\psi(i)})= \rho\left ( \begin{array}{cc} \color{blue}{i} & \color{blue}{0} \\ \color{blue}{0} & \color{blue}{-i} \end{array}\right)$$ Then $$ I = \left ( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array} \right )$$

so that

$$ I^2 = -1$$

Edit 3

After the discussion with Incnis Mrsi I calculated $J^2=I^2 =-1$ for $J=\rho_{2}(\psi_1(j))$ and $I=\rho_{2}(\psi_1(i))$. I am confused that this seems to work. The reason why I asked this quetion is the following passage in Tapp's matrix groups for undergraduates:

enter image description here

In particular,

why is $I\neq\rho_{2n}(\psi_n(i))$ and $J\neq\rho_{2n}(\psi_n(j))$ for $n>1$? (for $n=1$, apparently it works as I just verified).

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  • $\begingroup$ can you specify how are defined the homomorphisms $\rho_n$, $\psi_n$, $\rho_{2n}$ ? $\endgroup$ – Emilio Novati Dec 14 '14 at 19:15
  • $\begingroup$ Why the downvote? $\endgroup$ – learner Dec 27 '14 at 11:12
  • $\begingroup$ Not mine. I'm working on your question, but I don't well understand the problem. It seems to me that all is OK. Your problem is to find the matrices $I_{4n}$ and $J_{4n}$ ? $\endgroup$ – Emilio Novati Dec 29 '14 at 17:25
  • $\begingroup$ I've not the book, so in the commutative diagrams, what means $R_{J_{4n}}$ ? $\endgroup$ – Emilio Novati Dec 29 '14 at 17:43
  • $\begingroup$ $R_A$ is used to denote the linear map corresponding to the matrix $A$ (if given in the standard basis): $R_A(X) = (X^TA)^T$ where $^T$ denotes the transpose. The paragraph I included in the question gives $I_{4n}$ and $J_{4n}$ so I just want to know why $I_{4n}=\rho_{2n}(\psi_n(i))$ and $J_{4n}=\rho_{2n}(\psi_n(i))$ is not possible. $\endgroup$ – learner Dec 29 '14 at 23:46
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First of all, if your representation of quaternionic matrices gives $I^2≠−1$, it indicates you made a mistake and need to check which 4 × 4 real (or 2 × 2 complex) blocks do you use. You can consult this Wikipedia discussion for working examples (although there are many possible choices).

Second, your analogy between ℂ and ℍ is flawed because of poor algebraic design. Complex numbers form a field, that requires all element to commute, specifically $$∀z∈ℂ: i\,z = z\,i\,,$$ that implies said commutation requirement for matrix representations.

The algebra of quaternions isn’t a (true) field, it is only a skew field (a division algebra). $i\,z = z\,i$ isn’t an identity anymore, it’s an equation of ℂ (a two-dimensional real subalgebra of ℍ). Consequently, real (or complex) representation of quaternions, or quaternionic matrices, will not generally commute with any of $I, J, K$. You have to use other kind of matrix equation, such as ones using complex conjugate transpose (look again at the en.WP link for some insights).

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  • $\begingroup$ What you state in your first paragraph is wrong. Please see the calculations, I added them to my question. I thoroughly checked them and could find no mistake. $\endgroup$ – learner Dec 25 '14 at 2:31
  • $\begingroup$ @learner: really? I see no less than two. First, $i$ times identity matrix in Mat₂(ℂ) is $\begin{pmatrix}i&0\\0&i\end{pmatrix}$, not $\begin{pmatrix}0&i\\i&0\end{pmatrix}$. Second, neither of two results in $\begin{pmatrix}0&0&0&1\\0&0&-1&0\\0&-1&0&0\\1&0&0&0\end{pmatrix}$ in Mat₄(ℝ) for any reasonable choice of basis in ℂ²→ℝ⁴. $\endgroup$ – Incnis Mrsi Dec 25 '14 at 17:13
  • $\begingroup$ @learner: also, I suggest you to avoid statements like “what you state in your first paragraph is wrong” unless you are competent enough to understand and assess what does another party say and, preferably, point to apparent contradictions explicitly. IMHO here we don’t see such a case. $\endgroup$ – Incnis Mrsi Dec 25 '14 at 17:23
  • $\begingroup$ I said rubbish myself in the first comment. Of course, $I$ should not be $i$ times identity matrix, but $\begin{pmatrix}i&0\\0&-i\end{pmatrix}$ due to complex conjugation. $\endgroup$ – Incnis Mrsi Dec 25 '14 at 17:41
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    $\begingroup$ $ψ(i) = \begin{pmatrix}i&0\\0&-i\end{pmatrix}$ follows from $i = i\cdot 1 + 0\cdot j$. Full stop. $\endgroup$ – Incnis Mrsi Dec 26 '14 at 7:22
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Since I've not the book cited in OP and I'm not sure to well understand the notation that was used, I begin by fixing notation. We have two homomorphisms: $$ \rho: \mathbb{C} \rightarrow M_2(\mathbb{R}) \qquad \rho(z)=\rho(a+ib)= \left[ \begin{array}{ccccc} a&b \\ -b &a \end{array} \right] $$ and: $$ \psi: \mathbb{H} \rightarrow M_2(\mathbb{C}) \qquad \psi(x)=\psi(x_0+\mathbf{i}x_1+ \mathbf{j}x_2+\mathbf{k}x_3)= \left[ \begin{array}{ccccc} x_0+ix_1&x_2 +ix_3 \\ -x_2+ix_3 &x_0-ix_1 \end{array} \right] = X_{(2)} $$ that are homomorphisms between rings, so that: $$ \rho(wz)=\rho(w)\rho(z) \qquad \psi(xy)=\psi(x)\psi(y) $$ (Note that in second case the order is important because $\mathbb{H}$ is a noncommutative ring.)

We can genaralize $\rho$ as: $$ \rho_2: M_2(\mathbb{C}) \rightarrow M_4(\mathbb{R}) $$ $$ \rho_2(A) = \rho_2 \left( \left[ \begin{array}{ccccc} z_{1,1}&z_{1,2} \\ z_{2,1} &z_{2,2} \end{array} \right] \right) = \left[ \begin{array}{ccccc} \rho(z_{1,1})&\rho(z_{1,2}) \\ \rho( z_{2,1}) &\rho(z_{2,2}) \end{array} \right] $$ and more general: $$ \rho_n: M_n(\mathbb{C}) \rightarrow M_{2n}(\mathbb{R}) $$ $$ \rho_n(A) = \rho_n \left( \left[ \begin{array}{ccccc} z_{1,1}& \cdots &z_{1,n} \\ \cdots \\ z_{n,1} & \cdots &z_{n,n} \end{array} \right] \right) =\left[ \begin{array}{ccccc} \rho(z_{1,1})&\cdots &\rho(z_{1,n}) \\ \cdots \\ \rho( z_{n,1})&\cdots &\rho(z_{2,n}) \end{array} \right] $$ and analogously we can do for $\psi$ to define $\psi_n$:

Now, working in n=1 for simplicity, we have the homomorphisms: $$ \rho_2 \circ \psi : \mathbb{H} \rightarrow M_4(\mathbb{R}) $$ so defined: $$ \begin{split} \rho_2 \circ \psi(x)=\rho_2(\psi(x))=\rho_2(X_{(2)}) &= \\ = \left[ \begin{array}{ccccc} \rho(x_0+ix_1)&\rho(x_2 +ix_3) \\ \rho( -x_2+ix_3) &\rho(x_0-ix_1) \end{array} \right] &=\\ = \left[ \begin{array}{ccccc} x_0&x_1&x_2 &x_3 \\ -x_1&x_0&-x_3 &x_2 \\ -x_2&x_3&x_0 &-x_1 \\ -x_3&-x_2&x_1 &x_0 \end{array} \right] = X_{(4)} \end{split} $$ note that the rows of this matrix are the ordered quadruples of the components of the quaternionic vector $$ \left[ \begin{array}{ccccc} 1x \\ \mathbf{i}x\\ \mathbf{j}x \\ \mathbf{k}x \end{array} \right] $$ so we adopt the useful notation: $$ \rho_2 \circ \psi(x)= \left[ \left[ \begin{array}{ccccc} 1x \\ \mathbf{i}x\\ \mathbf{j}x \\ \mathbf{k}x \end{array} \right] \right] $$ now it's easy to see that $$ \rho_2 \circ \psi(\mathbf{i})=I_{(4)}= \left[ \left[ \begin{array}{ccccc} \mathbf{i} \\ -1\\ -\mathbf{k} \\ \mathbf{j} \end{array} \right] \right] = \left[ \begin{array}{ccccc} 0&1&0 &0 \\ -1&0&0 &0 \\ 0&0&0 &-1 \\ 0&0&1 &0 \end{array} \right] $$ as in OP and, in the same way, we can find $J_{(4)}=\rho_2 \circ \psi(\mathbf{j})$ and $K_{(4)}=\rho_2 \circ \psi(\mathbf{k})$.

And, more important, we easily see that $$ \rho_2 \circ \psi(\mathbf{i}x)= \left[ \left[ \begin{array}{ccccc} 1\mathbf{i}x \\ \mathbf{i}\mathbf{i}x\\ \mathbf{j}\mathbf{i}x \\ \mathbf{k}\mathbf{i}x \end{array} \right] \right]= \left[ \left[ \begin{array}{ccccc} \mathbf{i}x \\ -x\\ -\mathbf{k}x \\ \mathbf{j}x \end{array} \right] \right] $$ is different from $$ \mathbf{i} \left(\rho_2 \circ \psi(x)\right)= \mathbf{i} \left[ \left[ \begin{array}{ccccc} 1x \\ \mathbf{i}x\\ \mathbf{j}x \\ \mathbf{k}x \end{array} \right] \right]= \left[ \left[ \begin{array}{ccccc} \mathbf{i}x \\ -x\\ \mathbf{k}x \\ -\mathbf{j}x \end{array} \right] \right] $$ but we have: $$ \rho_2 \circ \psi(\mathbf{i}x)=\rho_2\left[\psi(\mathbf{i})\psi(x)\right]=\rho_2\left(\psi(\mathbf{i})\right)\rho_2\left(\psi(x)\right) $$ so that the diagram in OP commutes if we take, on the right side, $I_{(4)}=\rho_2\left(\psi(\mathbf{i})\right)$. I think that, with a bit of work, we can extend this reasoning to the case $n>1$.

I hope this can be an answer to the question. I'm not sure in my understanding of the page of the book, but I suspect that there is a mistake in the matrices named $\mathcal{I}_4$ and $\mathcal{J}_4$.

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