Embedding Complex Matrices into Real Spaces

[Kreizhn]

Hey all,

I have a quick question that should hopefully be simple to answer.

Consider a the space of [itex] n \times n [/itex] matrices over [itex] \mathbb C[/itex] given by [itex] M_n(\mathbb C) [/itex]. In order to properly consider this as a real matrix, we have to embed [itex] M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex], and I can give some books that cite this. In order to do this, we use one of the following methods:

Method 1
Define the map [itex] \rho: \mathbb C \to M_2(\mathbb R) [/itex] by
[tex] \rho(x+iy) = \begin{pmatrix} x & -y \\ y & x \end{pmatrix} [/tex]
which actually gives an identification of the complex numbers with a subspace of [itex] M_2(\mathbb R) [/itex]. Let [itex] Z \in M_n(\mathbb C) [/itex] be given by the components [itex] Z=[Z]_{ij}[/itex]. Then define [itex] \rho_n: M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex] as [itex] \rho_n(Z) = [\rho(Z_{ij})]_{ij} [/itex]. That is, we have just made each complex entry of Z into a 2x2 matrix.

Method 2

Similar to above, let [itex] Z = X+iY \in M_N(\mathbb C) [/itex] where [itex] X,Y \in M_n(\mathbb R) [/itex] and define [itex] \rho': M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex] by
[tex] \rho'(Z) = \begin{pmatrix} X & -Y \\ Y & X \end{pmatrix} [/tex]
where this is done in block-matrix form.

Now both of these methods are related, and the reason for the relation comes from the two different definitions of symplectic structures. However, this is not what I'm interested in. We notice that [itex] M_{2n}(\mathbb R) [/itex] is a [itex] 4n^2 [/itex] dimensional space and in particular we can identify it with [itex]\mathbb R^{4n^2} [/itex]. What if instead of using one of these embeddings, we instead wrote [itex] Z = X + i Y [/itex] and visualized it as an element of [itex] M_{n,2n}(\mathbb R) [/itex] with
[tex] Z = \begin{pmatrix} X \\ Y \end{pmatrix} [/tex]
which has dimension [itex] 2n^2 [/itex] and so we identify it with [itex] \mathbb R^{2n^2} [/itex]. Why do we prefer using one of the previous two methods if they give a higher dimensional space? Is it for some reason like "multiplication is properly preserved" or something along those lines?

Edit: Something in my head says that from a topological point of view, there's probably not a big difference. However, it would be something like "We want to preserve the ring structure of [itex] \mathbb C [/itex]."

 

[Citan Uzuki]

As you suspect, it is to preserve multiplication. After all, you can't even multiply two n×2n matrices together, so the mapping you suggest certainly won't preserve the product of the two matrices. So to get a ring homomorphism, we have to take [itex]\mathbf{M}_{n}(\mathbb{C}) \rightarrow \mathbf{M}_{2n}(\mathbb{R})[/itex]

 

[Kreizhn]

Would it not be possible to "vectorize" the [itex] n \times 2n [/itex] matrices and define some sort of really weird multiplication on [itex] \mathbb R^{2n^2} [/itex] that would preserve the ring structure?

 

[Citan Uzuki]

Sure you could, just take the injective linear map from [itex]\mathbf{M}_{n}(\mathbb{C}) \rightarrow \mathbb{R}^{2n^2}[/itex] and transfer the multiplication from [itex]\mathbf{M}_{n}(\mathbb{C})[/itex] to [itex]\mathbb{R}^{2n^2}[/itex]. However, having to define such nonstandard multiplication defeats the purpose of finding real matrix representations (which is to embed them in the real matrix ring with its usual multiplication).

 

[AlephZero]

Hey all,
Why do we prefer using one of the previous two methods if they give a higher dimensional space? Is it for some reason like "multiplication is properly preserved" or something along those lines?

From a computational point of view, that is back to front. The advantage of transforming from real variable to complex (when it is possible) is to reduce the matrix dimensions, and also reduce the amount of computation by doing everything only once, rather than twice.

If you are going to define something that doesn't actually behave like a matrix, then it's not a matrix. (If something doesn't look like a duck or quack like a duck, then it's not a duck).

The "double sized" real formulation arises naturally in some physical problems, for example in the dynamics of an object with "cyclic symmetry" like a fan with a number of identical equally spaced blades. You can represent the motion of the complete fan as a Fourier series of the degrees of freedom of one blade. If the fan is in the XY plane with its axis along z = 0, by symmetry the equatons of motion of the complete fan are the same in X and Y directions if you replace X by Y and Y by -X.

Putting those two facts together is equivalent to the matrix transformation between n complex variables and 2n real ones.

 

[Kreizhn]

So if I get what you're both saying, it's essentially that there is no theoretical issue with embedding [itex] M_n(\mathbb C) [/itex] into [itex] \mathbb R^{2n^2} [/itex]. However, for the purpose of keeping matrices as matrices, the only way to maintain the "matrix" structure is the transformation into [itex] M_{2n}(\mathbb R) [/itex].

Edit: Keep in mind that the quotation marks around "matrix" structure is meant to be entirely colloquial. I figure someone might get pedantic and say something like "the matrix structure and the ring structure are the same" so I just wanted to avoid that.