Rank Jacobian matrix product I take  $m, n, i \in \mathbb{N}^*$, with $i \leq m,n$. I consider the application $\phi:\mathbb{R}^{m \times i} \times \mathbb{R}^{i \times n} \to \mathbb{R}^{m \times n}$, with $\phi(W_1,W_2) = W_1W_2$. 
In an article, it is said that the rank of the Jacobian matrix (of size $mn \times (mi+ni)$) of $\phi $, $J_{\phi}(W_1,W_2)$ is equal to $in+im-i^2$ if $W_1$ and $W_2$ are of full rank, and strictly less if one of $W_1$ and $W_2$ is not of full rank. 
I did a proof of the formula when $W_1$ and $W_2$ are of full rank (by applying the Grassman formula to the linear space $\{W_1 U + V W_2, U \in \mathbb{R}^{i \times n}, V \in \mathbb{R}^{m \times i} \}$). My first question is: does someone know a reference where I could check this, and possibly find a shorter way.
My second question is: how to prove that the rank of the Jacobian matrix is strictly less than $im+in -i^2$ when one of $W_1$ and $W_2$ is not of full rank. 
Thanks!
 A: In order to solve your problem, we need to compute the dimension of the following subspace of $M_{m\times n}$:   $\{W_1U+VW_2;\ U\in M_{i\times n},\ V\in M_{m\times i}  \}$, where $W_1\in M_{m\times i}$ and $W_2\in M_{i\times n}$.
Let us prove the following proposition:  

Proposition. Let $i\leq\min\{m,n\}$. Let $p_1$ and $p_2$ be the ranks of $W_1\in M_{m\times i}$ and $W_2\in M_{i\times n}$
respectively, then $$\dim\ \{W_1U+VW_2;\ U\in M_{i\times n},\ V\in M_{m\times i}  \}= np_1+mp_2-p_1p_2.$$  

Now, if $p_1<i$ then $p_1(n-i)<i(n-i)$ then $np_1+mi-p_1i< ni+mi-i^2$. 
But, $np_1+(m-p_1)p_2\leq np_1+(m-p_1)i$ then $$np_1+mp_2-p_1p_2< ni+mi-i^2$$
You can do the same if $p_2<i$, just change $n$ by $m$.
Proof of Proposition:
Since rank of $W_i$ is $p_i$, we can write $W_1=L_1M_1$ $(L_1\in M_{m\times p_1}$, $M_1\in M_{p_1 \times i})$ and $W_2=M_2L_2$ $( M_2\in M_{i\times p_2}$, $L_2\in M_{p_2 \times n})$ and the ranks of $L_1$ and $M_1$ are equal to $p_1$ and the ranks of $L_2$ and $M_2$ are equal to $p_2$.
$($Notice that $L_1,M_2$ have left inverses and $M_1,L_2$  have right inverses$)$

(1) Now, prove that $$\{W_1U+VW_2;\ U\in M_{i\times n},\ V\in M_{m\times i}\}  =\{L_1R+SL_2;\ R\in M_{p_1\times n},\ S\in M_{m\times p_2}  \}.$$
Hint: $M_1$ has a right inverse $(M_1\overline{M_1}=Id_{p_1\times p_1})$ and $M_2$ has a left inverse $(\overline{M_2}M_2=Id_{p_2\times p_2})$.

So we need to compute the dimension of $\{L_1R+SL_2;\ R\in M_{p_1\times n},\ S\in M_{m\times p_2}  \}.$ 
We can compute this dimension using the Grassman formula that you said.
Since $L_1$ has a left inverse, the $\dim\{L_1R;\ R\in M_{p_1\times n}\}=np_1$. Since $L_2$ has a right inverse then   $\dim\{SL_2;\ S\in M_{m\times p_2}\}=mp_2$.
Let $\overline{L_1}$ be the left inverse of $L_1$ $(\overline{L_1}L_1=Id_{p_1\times p_1})$.
Now, $L_1R=SL_2$ implies that $R=\overline{L_1}SL_2$ and $L_1R=L_1\overline{L_1}SL_2$.
Notice that $\{L_1R;\ R\in M_{p_1\times n}\}\cap \{SL_2;\ S\in M_{m\times p_2}\}=\{L_1\overline{L_1}SL_2;\ S\in M_{m\times p_2}\}$.

(2) Prove that $\{L_1\overline{L_1}SL_2;\ S\in M_{m\times p_2}\}=\{L_1ZL_2;\ Z\in M_{p_1\times p_2}\}$.
Hint: $\overline{L_1}L_1=Id_{p_1\times p_1}$

Finally, since $L_1$ has a left inverse and $L_2$ a right inverse $\dim\{L_1ZL_2;\ Z\in M_{p_1\times p_2}\}=p_1p_2$.
So $\dim\{W_1U+VW_2;\ U\in M_{i\times n},\ V\in M_{m\times i}\}=np_1+mp_2-p_1p_2$.
 $\square$
