Inequalities of singular values 
My attempt: Let $A=U\Sigma V^*$ be the singular value decomposition of $A$ where $U$ and $V$ are unitary. Now let $\hat{A}=\begin{bmatrix}0 &A \\A^*& 0\\\end{bmatrix}$ and $\hat{\Sigma}=\begin{bmatrix}0&\Sigma\\\Sigma^*&0\\\end{bmatrix}$. From here what I'm trying to do is relate the eigenvalues of $\hat{A}$ to the eigenvalues of $\hat{\Sigma}$, but I can't see any relation. Is my approach right or I'm missing something? Any help would be much appreciated.
 A: Your approach a correct approach to prove the first inequality. Note that $\hat A$ is unitarily similar to $\hat\Sigma$, and you can show that the non-zero eigenvalues of $\hat \Sigma$ are equal to the non-zero eigenvalues of $\Sigma$, which are in turn the singular values of $\hat A$. From there, applying Weyl's inequalities to $\hat A, \hat B, \hat A + \hat B$ produces the desired result.

The below is my work in progress for the second part.
The second result is tricky to prove. One approach is as follows. First of all, we note that if either $A$ or $B$ have rank less than $i+j-1$, then the same holds for $AB^*$, which means that
$$
\sigma_{(i-1) + j}(AB^*) =  0,
$$
so that the inequality trivially holds
Now, suppose that $A$ and $B$ both have rank at least $i + j - 1$. We use the following notation:

*

*$u_j,\dots,u_q$ denote the left singular vectors of $B$ corresponding to $\sigma_j(B),\dots,\sigma_q(B)$.

*$U_1 = \operatorname{span}(\{u_j,\dots,u_q\}) \subset \Bbb C^m$

*$v_i \dots,v_q$ denote the right singular vectors of $A$ corresponding to $\sigma_i(A),\dots,\sigma_q(A)$

*$V_2 = \operatorname{span}(\{v_i,\dots,v_q\}) \subset \Bbb C^n$

*$U_2 = B^{-*}(V_2) \subset\Bbb C^m$.

If $\Pi_2$ denotes the projection onto $V_2^\perp = \operatorname{im}(A)$, then $U_2 = \ker(\Pi_2 \circ B^*)$, which has dimension at least $\min\{\operatorname{rank}(A),\dim \ker(B^*)\}$.
Let $r = \operatorname{rank}(B)$.
Let $v_1,\dots,v_j \in \Bbb C^m$ denote the leading left singular vectors of $B$, and let $V = \operatorname{span}(\{v_1,\dots,v_j\})$. For all vectors $x \in V$, it holds that $\|B^*x\| \geq \sigma_j(B)\|x\|$. Similarly, let $u_i,\dots,u_q$ denote the trailing right singular vectors of $A$ and take $U = \operatorname{span}(\{u_i,\dots,u_q\})$.
to use the min max theorem for singular values. Note that
$$
\sigma_{i+j-1}(AB^*) = \max_{\dim(U) = i+j} \min\{\|AB^*x\| :x \in U, \|x\| = 1\}
$$
