I have to proof the following statement:
Prove that for a given $A\in \mathcal M _{n \times m}(\mathbb R)$ there exist two orthogonal matrices $U \in \mathcal O(n)$, $V \in \mathcal O(m)$ such that: $$UAV^T=\begin{pmatrix}||A||_2 & 0\\ 0 & A_1 \end{pmatrix},$$ where $||A||_2$ is the 2 norm of the matrix and $A_1\in \mathcal M _{n-1 \times m-1}(\mathbb R)$.
The professor told me to use Gram-Schmidt orthonormalization process as a hint. My try so far has been the following.
I need to find two rotation matrices that rewrite $A$ in the way above. To find the first one, I've thought of generating a set of orthonormal vectors $\{v_1, v_2, ..., v_m\}$ where $v_1\in \mathbb R^m$ is a vector that fulfills $||Av_1||_2=||A||_2$. To find the second matrix, I've thought of using another rotation matrix that changes from the canonic basis in $\mathbb R^n$ to a orthogonal basis containing $\frac{Av_1}{||Av_1||_2}=\frac{Av_1}{||A||_2}$ as one of the vectors.
This way, I have two orthogonal matrices that applyed to the original matrix give the first column as I want it to be. However, I can't find a way to prove that the rest of the first row has to be al zeros other than the first element.
I think the solution has something to do with which linearly independent vectors $\{x_1,..., x_m\}$ I choose to perform Gram-Schmidt. If i could somehow find a way to find those vectors sutch that $Av_1 \perp \{Av_2, ..., Av_m\}$ id then have that the rest of the first row are zeros too.
Answering my own question here, apparently, if you choose the $U$ and $V$ matrices as the chamge of basis matrices from the canonic basis and two specific basis in $\mathbb R^m$ and $\mathbb R^n$ you can then prove that the first column is also zeros.
these two basis are $B_m=\{v_1,...,v_m\}$ and $B_n=\{u_1,...,u_m\}$ where $v_1 \in \mathbb R^m$ is the vector sutch that $||Av_1||=||A||$ and $w_1 \in \mathbb R^n$ is the vector sutch that $w_1=\frac{Av_1}{||A||}$.
This way, $M:=UAV^t$ is of the form: \begin{pmatrix} ||A|| & z^t \\ 0 & A_1 \end{pmatrix}
Now given $y \in \mathbb R^m$ of the form
\begin{pmatrix} ||A|| \\ z \end{pmatrix}
we can see that $||My|| \leq (||A||^2+||z||^2)^2$ and $||y||=(||A||^2+||z||^2)$ therefore, as the change of basis matrices are orthogonal,
$||A||=||M|| \leq ||My||/||y|| \leq (||A||^2+||z||^2)$ and therefore $||z||=0$ and $z=0$