# How to solve the inverse problem of least-squares?

Focusing on following least squares problem:

$$\min\limits_{V} \lVert Z - WV \rVert _{_F}^2$$

$$Z∈{R}^{m\times n},\quad W∈{R}^{m\times k},\quad V∈{R}^{k\times n},\quad k\lt m\lt n$$

This problem can be easily solved given $Z$ and $W$. I assume the solution of this problem as $\overline{V}$.

My question is that, how can we recover $W$ by $\overline{V}$ and $Z$?

I found something called inverse optimization, but cannot figure out the relationship between them.

The solution to the initial least squares problem can be written in terms of the pseudoinverse $${\bar V} = W^+Z$$ Then several days later, you still have $({\bar V},Z)$ but have somehow lost the definition of $W$.
You can try the following \eqalign{ W^+ &= {\bar V}Z^+ \cr W &= ({\bar V}Z^+)^+ \cr }
• Thanks for your answer. In your answer you assume $${W^+} = W^+ZZ^+$$ I am not sure if it is true. – lee Jun 6 '18 at 1:18
• It'll be true so long as $\,{\rm rank}(Z) = m\,\,$ – greg Jun 6 '18 at 4:06
• Thank you. @greg Your answer is very valuable. What if rank($Z$) <$m$? – lee Jun 6 '18 at 9:44
• @greg are you sure? $(e_1)^+ = e_1^T$, but $w^T = w^Te_1e_1^T$ does not hold for all vectors $w$. – Federico Poloni Jun 6 '18 at 16:20
• @FedericoPoloni The question stipulates that $Z\in{\mathbb R}^{m\times n}$ and $m<n$. Your counter-example is $e_1\in{\mathbb R}^{m\times 1}$, which does not satisfy the second criteria since $m>1\,\,$ – greg Jun 6 '18 at 18:11