# Shorter (more concise) writing of a certain block matrix

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.

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.