We consider matrix $A_{i, j} = \sin(i) \cdot \cos(j)$, $0 \le i, j < n$. It is needed to find the compact SVD for this matrix.
I think the rank is always 1 (for $n > 1$) since each row of this matrix is proportional to the second row but it is not easy to see what is the SVD for this matrix. I understand that for compact SVD I only need to find the first column of $U$ and the first column of $V$. Still don't understand how to find them.
They are already given to you. Just take $$a=\sqrt{\sum_{i=0}^{n-1}\sin^2i},\qquad\qquad b=\sqrt{\sum_{j=0}^{n-1}\cos^2j},$$ $$ U=\frac1a\begin{bmatrix}\sin 0\\ \vdots\\ \sin(n -1)\end{bmatrix},\qquad\qquad V=\frac1b\begin{bmatrix}\cos 0\\ \vdots\\ \cos(n -1)\end{bmatrix},\qquad\qquad\Sigma=ab. $$