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Given a commutative $*$-ring $(R,*)$, let $M(R,*)$ be the monoid whose elements are matrices over $R$ of all possible shapes and entries, including those that have $0$ columns or $0$ rows. Let the monoid operation be $\oplus$, denoting direct sum. Let $M \sim K$ be an equivalence relation expressing that there exist unitary matrices $U$ and $V$ such that $UMV = K$. The equivalence relation $\sim$ is therefore unitary equivalence. Has the monoid $M(R,*)/\sim$ ever been studied in the literature?

Note 1: The monoid $M(R,*) / \sim$ is commutative, while $M(R,*)$ is not commutative unless $R = 0$ (the zero ring).

Note 2: For $$(R,*) \in \{(0, \operatorname{id}_{0}), (\mathbb R, \operatorname{id}_\mathbb R), (\mathbb C, a + bi \mapsto a - bi), (\mathbb R[\varepsilon]/(\varepsilon^2),a+b\varepsilon \mapsto a+b\varepsilon), (\mathbb R[\varepsilon]/(\varepsilon^2),a+b\varepsilon \mapsto a-b\varepsilon)\}$$ we have that $M(R,*)/\sim$ is isomorphic to a free commutative monoid. In fact, I don't know of any $M(R, *)/\sim$ which is not isomorphic to a free commutative monoid. When such an isomorphism exists, it effectively generalises the Singular Value Decomposition. For the example $M(0, \operatorname{id}_0) / \sim$, the free commutative monoid has only two generators: the unique possible $0\times 1$ matrix and the unique possible $1 \times 0$ matrix.

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  • $\begingroup$ I’m pretty sure I remember this question being asked in the past but I couldn't find it searching $\endgroup$ Nov 10, 2021 at 10:40
  • $\begingroup$ @BenjaminSteinberg Probably by me, especially if it was asked recently. I got it wrong a few times. In my first formulation, I didn't consider any matrices with zero rows or zero columns except for the (obviously unique) $0 \times 0$ matrix. I realise now that I need to consider $0 \times n$ and $n \times 0$ matrices as well. In the case where $(R,*) = (\mathbb C, a + bi \mapsto a - bi)$, we have that the generators of $M(R,*)$ are the $1 \times 1$ matrices with positive entries as well as the unique $0 \times 1$ and $1 \times 0$ matrices. $\endgroup$
    – wlad
    Nov 10, 2021 at 10:43
  • $\begingroup$ The $0 \times 1$ and $1 \times 0$ generators allow for non-square and non-invertible matrices to have SVDs. The multiset of singular values can include those 2 matrices as well. The matrix $(0)$ on the other hand needs to be excluded from the generators. $\endgroup$
    – wlad
    Nov 10, 2021 at 10:47
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    $\begingroup$ @BenjaminSteinberg A $0 \times n$ matrix represents a linear map from $R^0$ to $R^n$. There is only one such map. For each $n$, the map is unique, but for different $n$ they are different. $\endgroup$
    – wlad
    Nov 10, 2021 at 19:12
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    $\begingroup$ @BenjaminSteinberg Yes. Non-square matrices are allowed $\endgroup$
    – wlad
    Nov 10, 2021 at 19:12

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