Proof of inequality for singular values I have two questions:
I wonder how to prove inequalities involving the singular values of matrix A*B such as
 
Using the definition meaning that largest singular value is max ||AX|| over all ||X||=1 and minimum singular value is min ||AX|| over all ||X||=1
 A: For matrices $A$, $B$, the singular values of $AB$ are the eigenvalues of
$$
       \sqrt{(AB)^{\star}(AB)} = \sqrt{B^{\star}A^{\star}AB}
$$
Using inner product notation: $\lambda_{A}$ is the smallest eigenvalue of $\sqrt{A^{\star}A}$ and $\Lambda_{A}$ its largest iff the following holds for all vectors $x$:
$$
              \lambda_{A}^{2}(x,x) \le (A^{\star}Ax,x) \le \Lambda_{A}^{2}(x,x).
$$
Building on this, let $x=By$:
$$
              \lambda_{A}^{2}(By,By) \le (A^{\star}ABy,By) \le \Lambda_{A}^{2}(By,By),\\
              \lambda_{A}^{2}(B^{\star}By,y) \le ((AB)^{\star}(AB)y,y) \le \Lambda_{A}^{2}(B^{\star}By,y)  \;\;\; (\dagger)
$$
The first inequality of $(\dagger)$ gives the following for all vectors $y$:
$$
                \lambda_{A}^{2}(B^{\star}By,y) \le  \Lambda_{AB}^{2}(y,y) \\
                   \implies \lambda_{A}^{2}\Lambda_{B}^{2} \le \Lambda_{AB}^{2}
$$
That's what you want. Similarly, the second of $(\dagger)$ gives $\Lambda_{AB}^{2} \le \Lambda_{A}^{2}\Lambda_{B}^{2}$.
