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

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,

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.

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 $$I_2 = \pmatrix{1&0\\0&1}, \quad J_2 = \pmatrix{0&-1\\1&0}$$

• Hi @Omnomnomnom, when you say that "suppose that A and B are similar.." you mean A and B complex, I think. And then phi(A) and phi(B) are again similar. In your second example, are you saying we can find A and B complex but not similar, for which phi(A) and phi(B) are similar? I think you might be missing some phi symbols, but I just want to make sure. And I think you mean phi(i) and phi(-i) are similar 4x4 real matrices (not 2x2). What do you think? It's a bit tricky for me to switch from a C-basis to an R-basis, so I just want to be sure I have it all correct. Thanks so much, – User001 Oct 16 '15 at 1:56
• You have most of that right. However, I really did mean 2x2 matrices, since $i$ and $-i$ are 1x1 – Omnomnomnom Oct 16 '15 at 13:08
• Ok, cool -- thanks so much @Omnomnomnom :-) – User001 Oct 17 '15 at 2:08
• @Omnomnomnom Why does $\phi$ have the homomorphism property? I.e. $\Phi(AB)=\Phi(A)\Phi(B)$. Is there an easy way to see this (other than explicitely calculating it)? – user527162 Dec 16 '18 at 14:25
• @mathsa the easiest approach I think is to use the properties of the Kronecker product, as I explain in the second part of my answer – Omnomnomnom Dec 16 '18 at 20:15