Representation of rank $d$ matrices with vectors

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.

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

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.