Given matrix $A\in R^{n\times m}$ prove that minimum of the $||A-xy^T||$, $||B||=tr(B^TB)$, is achieved when $x$ is an eigenvector of $AA^T$, corresponding to its greatest eigenvalue, and $y$ is an eigenvector of $A^TA$, corresponding to its greatest eigenvalue.
Progress I've made so far:
$f(x,y)=||A-xy^T||=tr(A^TA-2A^Txy^T+yx^Txy^T)$,
$\frac{\partial f(x,y)}{\partial x}=0 \Leftrightarrow tr(-2Ay+2xy^Ty)=0 \Leftrightarrow tr(xy^T)=tr(A)$
