Cond(AB) Vs Cond(A) and Cond(B) If a matrix G can be factorized as AB, A and B being non-square matrices, is there a way of expressing a relation between the singular value spectrum of G and the spectra of A and B ? Is there any bound that can be defined according to their respective conditioning numbers?
 A: EDIT . There is a mistake in my post. Then I completely rewrite it.
Let $U\in M_{p,n}$, $\sigma_1\geq \cdots\geq \sigma_k > 0,\cdots,0$ be its singular values and $U^+$ be its pseudoinverse. In a first time, we use the matricial norm $||.||_2$. Then $||U||=\sigma_1,||U^+||=1/\sigma_k$; we put, by definition, $cond(U)=||U||||U^+||$. Then $cond(U)=\sigma_1/\sigma_k$. 
Remark 1. Assume that $n\leq p$, $rank(U)=n$ and consider the least square problem $Ux=b$ where $b\in \mathbb{R}^p,x\in\mathbb{R}^n$; the solution is $x=U^+b=(U^TU)^{-1}U^Tb$. Then the accuracy of the calculation of $x$ depends on $cond(U^TU)$, that is $\sigma_1^2/\sigma_n^2=(cond(U))^2$. In other words, if $cond(U)=10^k$, then, when we calculate $x$, we lose $2k$ significant digits (and not only $k$).
Hypothesis 1.  (*) $A\in M_{np},B\in M_{pn}$ where $n\leq p$ and that $AB$ is invertible, that is, $A$ is onto and $B$ is injective; thus $A^T$ is injective, $im(B)\bigoplus \ker(A)=\mathbb{R}^p$ and $\ker(B^T)\cap im(A^T)=\{0\}$. Note that $A,B$ have $rank(A)=rank(B)=n$ non-zero singular values.
Proposition 1. Under (*), the inequality $cond(AB)\leq cond(A)cond(B)$ is not necessarily true.
Proof. One has $x^TABB^TA^Tx=(A^Tx)^T(BB^T)(A^Tx)\leq \rho(BB^T)x^T(AA^T)x\leq \rho(BB^T)\rho(AA^T)||x||^2$; then $\rho(ABB^TA^T)\leq \rho(BB^T)\rho(AA^T)$, that is $\sigma_1(AB)\leq \sigma_1(A)\sigma_1(B)$. 
Let $x\not= 0$; then $B^TA^Tx\not= 0$ and $(A^Tx)^T(BB^T)(A^Tx)\geq \sigma_n^2(B)x^TAA^Tx\geq \sigma_n^2(B)\sigma_n^2(A)x^Tx$, that implies $\sigma_n(AB)\geq \sigma_n(A)\sigma_n(B)$.
Unfortunately, the inequality concerning $\sigma_n$, is, in general, false; indeed $u^T(BB^T)u\not\geq \sigma_n^2(B)u^Tu$ (because $BB^T$ is not invertible and $u$ may have a component on $\ker(BB^T))$. That follows is a counter-example about the inequalities concerning $\sigma_n$ or $cond$.
Counter-example. Take $n=2,p=4$ and $A=\begin{pmatrix}99&-95.001&-25&76\\99&-95&-25&76\end{pmatrix},B=\begin{pmatrix}10&-62\\-44&-83\\26&9\\-3&88\end{pmatrix}$.
Remark 2. Under my above hypothesis $(*)$ (concerning $A,B$), in general, $(AB)^+\not= B^+A^+$. Yet, $(*)$ implies that $(BA)^+=A^+B^+$ and, in this case, the considered inequality is true as we'll see now . 
Hypothesis 2. (**) let $n\leq m,n\leq p$ and $A\in M_{m,n},B\in M_{n,p}$ s.t. $rank(A)=rank(B)=n$. The previous condition implies that $(AB)^+=B^+A^+$. 
Proposition 2. Under the hypothesis (**), the inequality $cond(AB)\leq cond(A)cond(B)$ is valid for any sub-multiplicative norm.
Proof (easy) $cond(AB)=||AB||||B^+A^+||\leq ||A||||B||||B^+||||A^+||=cond(A)cond(B).$
Remark 3. We see that the key, in the previous proof, is the relation $(AB)^+=B^+A^+$. There exists a NS condition:
Let $A,B\in M_n(\mathbb{C})$. Then $(AB)^+=B^+A^+$ IFF $BB^*(im(A^*))\subset im(A^*)$ AND if $K=im(A^*)\cap \ker(B^*)$, then  $A^*A(K)\subset K$.
Question. If the relation $(AB)^+=B^+A^+$ is not satisfied, then can we find a weaker condition that ensures that the inequality $cond(AB)\leq cond(A)cond(B)$ is true ?
