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Given a $n\times n$ matrix $U=[u_{ij}]$ and if we denote with ${\bf u}_k$ its columns ($n\times 1$ matrices), I wonder if there is a way to write the following $n^2\times n^2$ block matrix in an abbreviated/concise form, eventually using a Kronecker-like operation involving the matrix $U$, the vectorization of $U$, etc. This block matrix is

$$A=\left[\begin{array}{c|c|c} {\bf u}_1^T\otimes {\bf u}_1&\dots&{\bf u}_1^T\otimes {\bf u}_n\\ \hline \vdots&\ddots&\vdots\\ \hline \rule{0pt}{12pt}{\bf u}_n^T\otimes {\bf u}_1&\dots&{\bf u}_n^T\otimes {\bf u}_n\end{array}\right],$$ equivalent to $$A=\left[\begin{array}{c|c|c} {\bf u}_1\otimes {\bf u}_1^T&\dots&{\bf u}_n\otimes {\bf u}_1^T\\ \hline \vdots&\ddots&\vdots\\ \hline \rule{0pt}{12pt}{\bf u}_1\otimes {\bf u}_n^T&\dots&{\bf u}_n\otimes {\bf u}_n^T\end{array}\right]$$ or $$A=\left[\begin{array}{c|c|c} {\bf u}_1\cdot {\bf u}_1^T&\dots&{\bf u}_n\cdot {\bf u}_1^T\\ \hline \vdots&\ddots&\vdots\\ \hline \rule{0pt}{12pt}{\bf u}_1\cdot {\bf u}_n^T&\dots&{\bf u}_n\cdot {\bf u}_n^T\end{array}\right].$$

I have tried $\hbox{vec}(U)\cdot (\hbox{vec}(U))^T$ (with $\hbox{vec}$ being the column vectorization of a matrix), but it's not right.

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For a block matrix $$ M = \pmatrix{M_{11} & \cdots & M_{1n}\\ \vdots & \ddots & \vdots \\ M_{n1} & \cdots & M_{nn}}, $$ the partial transpose of $M$ (over the first space in the decomposition $\Bbb R^{n^2} = \Bbb R^n \otimes \Bbb R^n$) is given by $$ M^\Gamma = \pmatrix{M_{11} & \cdots & M_{n1}\\ \vdots & \ddots & \vdots \\ M_{1n} & \cdots & M_{nn}}. $$ You matrix can be expressed as $$ A = [\operatorname{vec}(U)\operatorname{vec}(U)^T]^\Gamma, $$ where vec denotes column-major vectorization.

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  • $\begingroup$ Thanks for the suggestion of using the partial transpose. $\endgroup$
    – JohnnyC
    Commented Mar 25, 2022 at 6:55

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