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Real $(n,n)$ (square) matrices are obviously a vector space. So we can consider linear transformations form this vector space to itself. Such a transformation $\mathsf{L}$ maps an $(n,n)$ matrix $\pmb{X}$ into another $(n,n)$ matrix, that is, $\mathsf{L}(\pmb{X})$ is an $(n,n)$ matrix and we also have $\mathsf{L}(a\pmb{X}+b\pmb{Y}) = a\, \mathsf{L}(\pmb{X}) + b\,\mathsf{L}(\pmb{Y})$ for every pair of $(n,n)$ matrices $\pmb{X}$ and $\pmb{Y}$, and every pair of real numbers $a$ and $b$.

What is the most general form of such a linear transformation, represented in terms of matrix multiplication and addition? I guess it must have the form $$ \mathsf{L} \colon \pmb{X} \mapsto \sum_{i=1}^{k} \pmb{A}_i\,\pmb{X}\,\pmb{B}_i $$ for some $(n,n)$ matrices $\pmb{A}_1, \dotsc, \pmb{A}_k$ and $\pmb{B}_1, \dotsc, \pmb{B}_k$.

  • Is this correct?

If so, then:

  • Are there general theorems that allow $\pmb{A}_i$ and $\pmb{B}_i$ to have specific properties a priori (eg, can they always be symmetric?); or that set a minimum value of $k$?

  • How are properties of the linear operator $\mathsf{L}$ – eg rank, determinant, symmetry or antisymmetry, eigensystem, transpose, inverse, and so on – reflected in the properties of $\pmb{A}_i$, $\pmb{B}_i$, $k$?

  • How does this representation and its properties generalize to affine transformations?

  • What are good references where to study this representation?

Thanks a bunch!

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    $\begingroup$ matrix multiplication is a "bilinear map" from $M\times M$ to $M$, not a linear one. Most general bilinear map is $\sum_{i,j }\alpha_{ij} A_{ij} B_{ij}$. Otherwise, it's not fully clear what you mean. $\endgroup$ Oct 15, 2020 at 7:54
  • $\begingroup$ @PeterFranek I rephrased title and question. Is it clearer now? $\endgroup$
    – pglpm
    Oct 15, 2020 at 7:56
  • $\begingroup$ I agree that the formula defines a linear map. I don't see what "such a linear operator, represented in terms of matrix multiplication? " means, so first part of the question (if it is correct) is not clear to me. For non-square matrices, you can do the same formula, there is no difference, just $A_i$ and $B_i$ will have other dimension. $\endgroup$ Oct 15, 2020 at 8:02
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    $\begingroup$ @PeterFranek I tried another rephrasing $\endgroup$
    – pglpm
    Oct 15, 2020 at 8:03

1 Answer 1

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Any linear map can be represented this way. Consider a matrix $A$ that has $1$ on position $i_1,j_1$ and zeros elsewhere, and matrix $B$ that has value $x$ on position $i_2,j_2$ and zero elsewhere. So then $AXB$ has value $x X_{j_1, \,i_2}$ on position $(i_2, j_1)$. So you can represent any linear map from matrices to matrices in this form -- at least if $k=n^4$.

If $A$ has shape $(m, n)$, it's still the same --- just $A_i$'s would have shape $(m,m)$ and $B_j$'s would have shape $(n,n)$.

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    $\begingroup$ Thank you. I suppose you mean $x\, X_{\dotso,\dotso}$ instead of $xA$. I wonder, though, if it's always possible to use some alternative representation, for example one that uses less than $n^4$ summands, or that uses symmetric matrices, and so on. $\endgroup$
    – pglpm
    Oct 15, 2020 at 8:21
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    $\begingroup$ Thank you again. I hope someone can provide at least some literature. I'm not sure about what you say on symmetry: even if $A,B,X$ are symmetric, the product $AXB$ doesn't need to be, unless $A=B$. $\endgroup$
    – pglpm
    Oct 15, 2020 at 8:36

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