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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

bsbb4
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Anonymous
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1 Answers1

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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.

greg
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