What does this notation mean? Linear map A = [A^\mu \nu]_\mu \nu

[nonequilibrium]

It's used in a certain proof that I'm reading. A is a linear map from a vectorspace V onto itself.

They say they can rewrite the vector space as $\mathcal V = \bigoplus_\mu \mathbb C^{m_\mu} \otimes \mathcal V^\mu$ and I understand this, but they then claim one can (always, as any linear map) rewrite A as $A = [A^{\mu \nu}]_{\mu \nu}$ "where $A^{\mu \nu}$ is a linear map of $\mathbb C^{m_\nu} \otimes \mathcal V^\nu$ to $\mathbb C^{m_\mu} \otimes \mathcal V^\mu$."

I don't understand the nature of this decomposition/rewriting. Note that this rewriting has to be possible for any A, it doesn't use any special properties of A (that comes later in the proof).

 

[morphism]

This seems to be an extremely obtuse way of writing the matrix of A in a given basis (indexed by \mu in the domain and by \nu in the range).

 

[nonequilibrium]

Writing a matrix in a given basis is possible, but they also fix the codomain for each part of the domain (it seems); I don't see why that is possible.

 

[morphism]

Do they explicitly state what they mean by $\mathcal V^\mu$?

 

[nonequilibrium]

Simply different vector spaces labelled by mu, it's as general as that.

 

[morphism]

Okay, so they are just writing A in terms of a given basis.

For simplicity, let's suppose $V = V^1 \oplus V^2$. Then, with respect to this decomposition, we can write
$$A = \begin{pmatrix} A^{11} & A^{12} \\ A^{21} & A^{22}\end{pmatrix}.$$
Here A^{11} will take V^1 to V^1, and A^{12} will take V^2 to V^1, etc.

 

[nonequilibrium]

Oh that makes sense, thank you :)