Hadamard product tricks of particular entities Consider the following matrices $\mathbf{Q}_{H}$ order $\left(
T\times n\right) $ and $\mathbf{A}$ of order $\left( T\times T\right) $ and $%
\mathbf{\hat{u}}$ of order $\left( n\times 1\right) $ and denote by $\circ $
the Hadamard product and by $\left\vert \mathbf{\hat{u}}\right\vert ^{\circ
-1}=\left\{ 1/\left\vert \hat{u}_{i}\right\vert \right\} $ (element wise
inverse of the vector $\mathbf{\hat{u}}$)
Observe that 
$$
\mathbf{A}\left( \mathbf{Q}_{H}\mathbf{\hat{u}}\circ \left\vert \mathbf{Q}%
_{H}\mathbf{\hat{u}}\right\vert ^{\circ -1}\right) \not=\mathbf{AQ}_{H}%
\mathbf{\hat{u}}\circ \left\vert \mathbf{Q}_{H}\mathbf{\hat{u}}\right\vert
^{\circ -1}
$$
thus
 \begin{eqnarray*}
\left\vert \mathbf{Q}_{H}\mathbf{\hat{u}}\right\vert \circ \left( \mathbf{A}%
\left( \mathbf{Q}_{H}\mathbf{\hat{u}}\circ \left\vert \mathbf{Q}_{H}\mathbf{%
\hat{u}}\right\vert ^{\circ -1}\right) \right)  &\not=&\left\vert \mathbf{Q}%
_{H}\mathbf{\hat{u}}\right\vert \circ \left( \mathbf{AQ}_{H}\mathbf{\hat{u}}%
\circ \left\vert \mathbf{Q}_{H}\mathbf{\hat{u}}\right\vert ^{\circ
-1}\right)  \\
&\not=&\left\vert \mathbf{Q}_{H}\mathbf{\hat{u}}\right\vert \circ \mathbf{AQ}%
_{H}\mathbf{\hat{u}}\circ \left\vert \mathbf{Q}_{H}\mathbf{\hat{u}}%
\right\vert ^{\circ -1} \\
&\not=&\mathbf{AQ}_{H}\mathbf{\hat{u}}
\end{eqnarray*}
Is there a mathematical trick to get rid of the $\left\vert \mathbf{Q}_{H}%
\mathbf{\hat{u}}\right\vert $ in this entity
$$
\left\vert \mathbf{Q}_{H}\mathbf{\hat{u}}\right\vert \circ \left( \mathbf{A}%
\left( \mathbf{Q}_{H}\mathbf{\hat{u}}\circ \left\vert \mathbf{Q}_{H}\mathbf{%
\hat{u}}\right\vert ^{\circ -1}\right) \right) 
$$
Or maybe another way of writing it that will help some of the proofs. 
Thank you so much in advance

Define the vectors 
$$v = Qu, \quad s={\rm sign}(v), \quad b={\rm abs}(v)$$
where the functions are applied elementwise.
Assuming $\,v_k\ne0$, the elementwise division of the vector is
$$v\circ b^{\circ -1} = v\oslash b = s$$
So the main equation reduces to
$$\eqalign{
y = b\circ(As) = BA\,s \cr
}$$
where the Hadamard product was eliminated by introducing the diagonal matrix
$$B = {\rm Diag}(b)$$
If some of the elements $\,v_k=0$, you can proceed by defining the corresponding $\,s_k=0$.
Another interesting way of writing the result is
$$\eqalign{
C &= bs^T, \quad
y &= (C\circ A)\,e \cr
}$$
where $e$ is the all ones vector.
