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In a paper that I am reading, it looks like the following property is used implicitly :

Let $M \in \mathbb{R}^{m \times n}$ a matrix of rank $d \leq \min(m,n).$ If $\left\{x_1,\dots,x_m\right\}$ is a base of $\mathbb{R}^{m}$ and $\left\{z_1,\dots,z_n\right\}$ is a base of $\mathbb{R}^{n}$, then there exists a matrix $ \Theta \in \mathbb{R}^{m \times n}$ of rank $d$ such that $M_{i,j}=x_i^T \Theta z_j$ for all $i,j$.

Is it true ? I do not understand where this comes from.

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    $\begingroup$ consider $\Theta := \big(X^{T}\big)^{-1}M Z^{-1}$ $\endgroup$ Commented Sep 21, 2023 at 15:58

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If the $x_i$ and the $z_j$ are the standard bases, we can choose $\Theta = M$, since $x_i^T \Theta z_j = \Theta_{ij}$, meaning every entry of $\Theta$ must agree with those of $M$. If the $x_i$ and $z_j$ are not the standard bases, we can choose change-of-basis matrices $T_{x}$ and $T_{z}$ such that $T_x x_i = e_i$ and $T_z z_j = e_j$ for all $i, j$. Then $T_x$ and $T_z$ will be invertible, and if we pick $\Theta = T_x^T M T_z$, we get \begin{align*}x_i^T \Theta z_j & = x_i^T T_x^T M T_z z_j \\ & = (T_x x_i)^T M T_z z_j \\ & = e_i^T M e_j \\ & = M_{ij}.\end{align*} Moreover , the rank of $\Theta = T_x^T M T_z$ and the rank of $M$ are equal since both $T_x^T$ and $T_z$ are invertible.

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