Let $M \in \mathbb{R}^{m \times n}$ be a matrix of rank $r$ with compact SVD $M=U\Sigma V^T$ ($U \in \mathbb{R}^{m \times r}$ and $V \in \mathbb{R}^{n \times r}$ are semi-unitary matrices and $\Sigma$ only contains the strictly positive spectral values of $M$). Let also $U_{\perp} \in \mathbb{R}^{m \times (m-r)}$ (resp. $V_{\perp} \in \mathbb{R}^{n \times (n-r)}$) be a matrix made of orthonormal vectors completing $U$ (resp $V$) in order to obtain an orthonormal basis of $\mathbb{R}^{m}$ (resp. $\mathbb{R}^{n}$).
It can be shown that for every $1\leq i \leq m$ and $1 \leq j \leq n $, $M_{i,j}=\phi(i,j)^T\theta$ where $\phi(i,j)$ and $\theta$ are vectors of size $d:=r(m+n)-r^2$ that are functions of $M$. Here are their precise definitions : $$\phi(i,j)=\begin{bmatrix} \operatorname{vec}(V_j^TU_i) \\ \operatorname{vec}(V_{\perp,j}^TU_i) \\ \operatorname{vec}(V_j^TU_{\perp,i}) \end{bmatrix}, \text{ } \theta=\begin{bmatrix} \operatorname{vec}(U^TMV) \\ \operatorname{vec}(U^TMV_{\perp}) \\ \operatorname{vec}(U_{\perp}^TMV) \end{bmatrix}, $$where $A_i$ is the $i$-th row of the matrix $A$ and $\operatorname{vec}$ is the vectorization operation.
My question is as follows. Is the following statement true or false : for any $\mu \in \mathbb{R}^{d}$, the matrix $M^{\mu} \in \mathbb{R}^{m \times n}$ defined by $M^{\mu}_{i,j}=\phi(i,j)^T\mu$ is of rank at most $r$.
With $M=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$ and $\mu=\left(\mu_1,\dots,\mu_8\right)$, one can show through tedious calculations that $M^{\mu}=\begin{pmatrix} \mu_1 & \mu_3 & \mu_7 \\ \mu_2 & \mu_4 & \mu_8 \\ \mu_5 & \mu_6 & 0 \\ \end{pmatrix}$ : it is easy to choose $\mu$ such that $M^{\mu}$ is full rank.