Notation for the direct sum and Kronecker sum of matrices

Question

The tensor product $w = u \otimes v$ of two vectors $u \in \mathbb{R}^m$ and $v \in \mathbb{R}^n$ is usually defined as \begin{equation} w_{in + j} = u_i v_j \text, \end{equation} and the Kronecker product of two (appropriately-sized) matrices $A \otimes B$ acts on such elementary tensors as \begin{equation} (A \otimes B) (u \otimes v) = (A u) \otimes (B u) \text. \end{equation}

The direct sum $w' = u \oplus v$ of two vectors is usually defined as \begin{equation} w'_i = \begin{cases} u_i & \text{if $i \le m$,} \\ v_{i - m} & \text{if $m < i \le n$,} \end{cases} \end{equation} and the direct sum of two matrices acts on this sum as \begin{equation} (A \oplus B) (u \oplus v) = (A u) \oplus (B u) \text. \end{equation}

One can also define the Kronecker sum of square matrices \begin{equation} A \mathop{\oplus_{\mathrm{Kronecker}}} B = A \otimes I_n + I_m \otimes B \text, \end{equation} such that \begin{equation} (A \mathop{\oplus_{\mathrm{Kronecker}}} B) (u \otimes v) = (Au) \otimes v + u \otimes (Bv) \text. \end{equation} This last operation is interesting in e.g. the theory of continous-time Markov chains, where if $A$ and $B$ represent infinitesimal generator matrices of two non-interacting systems, then $A \mathop{\oplus_{\mathrm{Kronecker}}} B$ is the infinitesimal generator for the (asynchronous) composition of the two systems.

Unfortunately, both the direct sum and the Kronecker sum of matrices is denoted by $\oplus$. What are some "standard" ways to handle this notational confusion if one has to work with mixed direct sums and Kronecker sums of matrices? The notation \begin{equation} A \mathop{\oplus_{\mathrm{Kronecker}}} (B \oplus C) \end{equation} is very cumbersome.

Wikipedia claims (unfortunately, I didn't find the source) that sometimes $\boxtimes$ is used to denote Kronecker products. Analogously, I could use $\boxplus$ for Kronekcer sums. However, I would prefer to use $\otimes$ and $\oplus$ for the Kronecker operators, mostly for aesthetic reasons (the box $\square$ is very "pointy").

Another way would be to use alternative notation for the direct sum. I guess by a not-so-slight abuse of notation, I could use $\times$ for the direct sum of vectors. $u \times v \in U \times V$ does not look so bad, and I don't need to work with infinte products where the direct sum and product are actually different. On the other hand, this feels like a huge hack...

Answer 1

I recently encountered a way to distinguish these that I thought was rather elegant:

direct product/sum - $A \otimes B$, $A \oplus B$

Kronecker product/sum - $A \hat{\otimes} B$, $A \hat{\oplus} B$

So, for example, $I_3 \hat{\otimes} B = B \oplus B \oplus B$ clearly shows this specific Kronecker product is expressed in terms of direct sums.

However, the hat on the Kronecker product is not strictly necessary, unless there is a desire for consistency with the way the Kronecker and direct sums are distinguished using this method.

Reference: Chirikjian, G. "Stochastic Models, Information Theory, and Lie Groups, Volume 1: Classical Results and Geometric Methods. Applied and Numerical Harmonic Analysis." (2009)

I found this reference on Google Books when searching for "kronecker sum" vs "direct sum". I had not seen this notation or anything like it in the dozen or so other references I've encountered, so it caught my attention. In most references, it is usually explained in the adjacent text which sum is meant. (Not that you asked, but to be complete I will mention that the products are less important to distinguish, since it is clear from the definitions of the symbols which product is meant. If the symbols represent matrices or vector spaces, then the symbol means Kronecker or direct product, respectively.)

Answer 2

I've seen people use "$A+B$" for the Kronecker sum. Since $A$ and $B$ act on different spaces and so can't be added normally, the reader is supposed to think: "I need to view $A$ and $B$ as acting on the same space. Aha! So '$A$' must mean '$A\otimes I$' and '$B$' must mean '$I\otimes B$' so that '$A+B$' refers to '$A\otimes I+I\otimes B$'".