Given that $A$ is $m\times n$ real matrices and then there exists orthogonal matrices $M_{m\times m}$ and $N_{n\times n}$ such that we get svd of $A=M\begin{bmatrix}\sigma_A&0\\0&0\end{bmatrix}N^T$ , now could anyone help me to understand how
$(1)$ $M^TA=\begin{bmatrix}\sigma_{\alpha}\\0\end{bmatrix}; \sigma_{\alpha} \text{ is }{\alpha\times n}$, a full row rank matrix.
$(2)$ $E$ is $q\times n$, $B$ is $q\times m$ real matrix, my question is does there exists any orthogonal matrix $U$ such that $U_{q\times q}[E\hspace{0.3cm} B]_{q\times (n+m)}=\begin{bmatrix}\bar{E}&\bar{B}\\0&0\end{bmatrix}_{q\times(n+m)}$ and $[\bar{E}\hspace{0.3cm} \bar{B}]_{?\times ?}$ of full row rank?
if $$A=M\begin{bmatrix}\sigma_A&0\\0&0\end{bmatrix}N^T$$
Since $M$ is orthogonal $M^TM=I$.
$$M^TA=\begin{bmatrix}\sigma_A&0\\0&0\end{bmatrix}N^T = \begin{bmatrix} \begin{bmatrix} \sigma_A & 0\end{bmatrix}N^T \\ \begin{bmatrix} 0 & 0\end{bmatrix}N^T \end{bmatrix}$$
Notice that the rows of $\begin{bmatrix} \sigma_A & 0\end{bmatrix}N^T$ consists of positive multiples of rows of $N^T$.
I will leave the second part for you.