Moore-Penrose inverse multiplication

Question

I really need help proving that when $AB=0$ then $B^+A^+=0$ and also the other way: when $B^+A^+=0$ then $AB=0$. Where $B^+$ and $A^+$ are Moore-Penrose Pseudo-inverse of B and A.

Answer 1

$$ \text{Let} \space A = U_1\Sigma_1V_1^* \space \text{and} \space B = U_2\Sigma_2V_2^* \space \text{(SVD of $A$ and $B$)} \\ \text{Then} \space A^+ = V_1\Sigma_1^+U_1^* \space \text{and} \space B^+ = V_2\Sigma_2^+U_2^* \\ $$

---

$$ \text{Now} \space AB = 0 \\ U_1\Sigma_1V_1^*U_2\Sigma_2V_2^* = 0\\ U_1^*U_1\Sigma_1V_1^*U_2\Sigma_2V_2^*V_2 = 0\\ \Sigma_1V_1^*U_2\Sigma_2 = 0 \\ \Sigma_1^+\Sigma_1^+\Sigma_1V_1^*U_2\Sigma_2\Sigma_2^+\Sigma_2^+ = 0 \\ \Sigma_1^+V_1^*U_2\Sigma_2^+ = 0 \\ \left(\Sigma_1^+V_1^*U_2\Sigma_2^+\right)^* = 0 \\ \Sigma_2^+U_2^*V_1\Sigma_1^+ = 0 \\ V_2\Sigma_2^+U_2^*V_1\Sigma_2^+U_1^* = 0 \\ B^+A^+ = 0 $$ You can show the converse by using the above proof and the fact that $\left(P^+\right)^+= P$ for any matrix $P$.