# On the use of generalized inverses for a particular case

I've the following question. Consider the equality $$A = C B D$$ with $B\in\mathbb{R}^{n\times n}$, $C\in\mathbb{R}^{k\times n}$ and $D\in\mathbb{R}^{n\times k}$. Particularly, $n>k$ and $C$ and $D$ have full rank. I would like to find an explicit expression of $B$ as a function of $A,C,D$.

How should I pick a generalized inverse properly?

Thanks in advance!

## 1 Answer

Whenever the equation is solvable, i.e. whenever $A=CB_0D$ for some $B_0$, you always have $B=C^+AD^+$ as a solution, because $$CBD=CC^+AD^+D=CC^+CB_0DD^+D=CB_0D=A.$$ In fact, by vectorising both sides of the equation $A=CBD$, we see that $$\operatorname{vec}(B)=(D^T\otimes C)^+\operatorname{vec}(A)=\left((D^+)^T\otimes C^+\right)\operatorname{vec}(A)=\operatorname{vec}(C^+AD^+)$$ is the least-norm solution to the least-square problem $\min_{B\in M_n(\mathbb R)}\|A-CDB\|_F^2$.

• Thanks for the quick response. Assume that I've one particular $B_0$ as you said, and I want a relationship of the type $B_0 = f(A,C,D)$ where $f$ is some matrix valued function of $A$ $C$ and $D$. Of course, such a function will be based on the relation $A = CB_0D$. Is that even possible? Thanks!!!! Commented Aug 12, 2017 at 15:50
• @NicoF. I don't quite understand your question. Anyway, if, say, $A_t,C_t,D_t$ are continuous functions in $t$ and $A_0=C_0B_0D_0,\ A_1=C_1B_1D_1$, there might not exist a continuous function $B_t=f(A_t,C_t,D_t)$ such that $A_t=C_tB_tD_t$ on $[0,1]$. Commented Aug 12, 2017 at 20:48