Are similar complex matrices again similar when each is expressed as a real matrix?

Question

We know that, relative to this ordered basis {$(1,0),(i,0),(0,1), (0,i)$}, we can express a 2x2 complex matrix mapping $C^2 -> C^2$ as a $4x4$ real matrix (representing the same transformation of $C^2$ into itself.)

Besides the usual, mechanical way of computing the matrix representations relative to a basis, if we use the specific basis mentioned above, then the change is simple, visually: each complex entry a+bi just gets replaced with a 2x2 real block:

$$ \begin{bmatrix} a & -b \\ b & a\\ \end{bmatrix} $$

My question is: some notes that I read about this change from complex entries to real 2x2 blocks state that this is an "operator isomorphism".

From this, can we tell whether two similar complex matrices are again similar, if we express each as a 4x4 real matrix? And the other way around: if two 4x4 real matrices are similar, then after expressing each in a 2x2 complex matrix, are these complex matrices again similar?

I know that I can compute the trace and determinants of the real and complex matrices and make some comparisons; for example, I computed the trace to be $tr(A_r)_{4x4}$ = 2 [$Re$ $tr(A_c)_{2x2}$].

But is there some intuitive explanation regarding the similarity / invertibility preservation between complex matrices and their corresponding real matrices?

Thanks,

Answer 1

Let $\Phi:\Bbb C^{n \times n}\to \Bbb R^{2n \times 2n}$ be the map that replaces each element $a+bi$ with the block $ \left(\begin{smallmatrix} a&-b\\b&a \end{smallmatrix}\right) $. This map has the following notable properties: For $A,B \in \Bbb C^{n \times n}$,

• $\Phi$ is $\Bbb R$-linear
• $\Phi(AB) = \Phi(A)\Phi(B)$
• $\Phi(A) = 0 \iff A = 0$
• $\Phi(A) = I \iff A = I$

Now, suppose that $A$ and $B$ are similar. That is, $A = SBS^{-1}$. It follows that $$ \Phi(A) = \Phi(SBS^{-1}) = \Phi(S)\Phi(B)\Phi(S)^{-1} $$ So, $\Phi(A)$ is similar to $\Phi(B)$.

On the other hand, things don't work so nicely in the other direction: we can find $A$ and $B$ that are not similar for which $\Phi(A)$ and $\Phi(B)$ are similar. For a $1 \times 1$ example, take $A = i$, $B = -i$. Note that $\Phi(i)$ and $\Phi(-i)$ are similar $2 \times 2$ matrices.

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There is a nice way to think about this map in terms of tensor and Kronecker products. In particular, we begin with a map $\Phi:\Bbb C^n \to \Bbb R^{2n} = \Bbb R \otimes \Bbb R^2$ of the form $$ \Phi(x + iy) = x \otimes e_1 + y \otimes e_2 $$ After seeing what $i$ does to a vector of this form, we can get a formula for arbitrary maps. In particular, we get $$ \Phi(A + Bi) = A \otimes I_2 + B \otimes J_2 $$ where $$ I_2 = \pmatrix{1&0\\0&1}, \quad J_2 = \pmatrix{0&-1\\1&0} $$