Need help understanding a paper's statement. Given
$$
\boldsymbol{E}_\ell= \alpha\left(\boldsymbol{I}+\alpha \boldsymbol{Z}_{\ell} \boldsymbol{Z}_{\ell}^{*}\right)^{-1}
$$
where $\boldsymbol{Z}_\ell \in \mathbb{R}^{n\times m}$ and $\boldsymbol{E}_\ell \in \mathbb{R}^{n\times n}$.
A paper claims that this is related to Ridge autoregression residual, and it states:
$$
\tag{1}
\boldsymbol{E}_{\ell} \boldsymbol{z}_{\ell}=\alpha\left(\boldsymbol{z}_{\ell}-\boldsymbol{Z}_{\ell}\left[\boldsymbol{q}_{\ell}\right]_{\star}\right)
$$
where $\left[\boldsymbol{q}_{\ell}\right]_{\star} = \underset{\boldsymbol{q}_{\ell}}{\operatorname{argmin}} \alpha\left\|\boldsymbol{z}_{\ell}-\boldsymbol{Z}_{\ell} \boldsymbol{q}_{\boldsymbol{\ell}}\right\|_{2}^{2}+\left\|\boldsymbol{q}_{\ell}\right\|_{2}^{2}$.
I know the formulation of $[\boldsymbol{q}_{\ell}]_{\star}$ ends up with the solution of Ridge regression.
And I know $\boldsymbol{z}_{\ell}-\boldsymbol{Z}_{\ell}\left[\boldsymbol{q}_{\ell}\right]_{\star}$ will then be the regression residual.
However, I cannot see the reason for equation (1) to hold. Any idea?
 A: \begin{align}
\nabla_{\boldsymbol q} \left(\alpha\left\|\boldsymbol{z}_{\ell}-\boldsymbol{Z}_{\ell} \boldsymbol{q}\right\|_{2}^{2}+\left\|\boldsymbol{q}\right\|_{2}^{2}\right) &= \alpha \nabla_{\boldsymbol q} \left(\boldsymbol z_{\ell}^*\boldsymbol z_{\ell} - 2\boldsymbol q^*\boldsymbol Z_{\ell}^* \boldsymbol z_{\ell}+\boldsymbol q^*\boldsymbol Z_{\ell}^*\boldsymbol Z_{\ell}\boldsymbol q\right) + \nabla_{\boldsymbol q}\left(\boldsymbol q^*\boldsymbol q\right)\\
&= \alpha \left(-2\boldsymbol Z_{\ell}^* \boldsymbol z_{\ell} + 2\boldsymbol Z_{\ell}^*\boldsymbol Z_{\ell}\boldsymbol q\right) +2 \boldsymbol q\\
&= -2\alpha \boldsymbol Z_{\ell}^*\boldsymbol z_{\ell} + 2\left(\boldsymbol{I}+\alpha \boldsymbol{Z}_{\ell}^*\boldsymbol{Z}_{\ell}\right)\boldsymbol q 
\end{align}
Since $\left[\boldsymbol{q}_{\ell}\right]_{\star} = \underset{\boldsymbol{q}_{\ell}}{\operatorname{argmin}} \alpha\left\|\boldsymbol{z}_{\ell}-\boldsymbol{Z}_{\ell} \boldsymbol{q}_{\boldsymbol{\ell}}\right\|_{2}^{2}+\left\|\boldsymbol{q}_{\ell}\right\|_{2}^{2}$, then $$-\alpha \boldsymbol Z_{\ell}^*\boldsymbol z_{\ell} + \left(\boldsymbol{I}+\alpha \boldsymbol{Z}_{\ell}^* \boldsymbol{Z}_{\ell}\right)\left[\boldsymbol{q}_{\ell}\right]_{\star} = 0 \implies \left(\boldsymbol{I}+\alpha \boldsymbol{Z}_{\ell}^* \boldsymbol{Z}_{\ell}\right)\left[\boldsymbol{q}_{\ell}\right]_{\star} = \alpha \boldsymbol Z_{\ell}^*\boldsymbol z_{\ell} $$
Let $\boldsymbol{y}_{\ell} = \alpha \left(\boldsymbol z_{\ell} -\boldsymbol{Z}_{\ell}\left[\boldsymbol{q}_{\ell}\right]_{\star}\right)$ it is clear that
\begin{align}
\frac1\alpha\left(\boldsymbol I + \alpha\boldsymbol Z_{\ell}\boldsymbol Z_{\ell}^{*}\right)\boldsymbol y_{\ell} &= \left(\boldsymbol I + \alpha\boldsymbol Z_{\ell}\boldsymbol Z_{\ell}^{*}\right)\left(\boldsymbol z_{\ell} -\boldsymbol{Z}_{\ell}\left[\boldsymbol{q}_{\ell}\right]_{\star}\right)\\
&= \boldsymbol z_{\ell} + \boldsymbol Z_{\ell} \left(\alpha \boldsymbol Z^*_{\ell}\boldsymbol z_{\ell}\right) - \boldsymbol Z_{\ell}\left[\boldsymbol{q}_{\ell}\right]_{\star} -\boldsymbol{Z}_{\ell}\left(\alpha \left(\boldsymbol{Z}_{\ell}^* \boldsymbol{Z}_{\ell}\right)\left[\boldsymbol{q}_{\ell}\right]_{\star}\right)\\
&= \boldsymbol z_{\ell} + \boldsymbol Z_{\ell} \left(\underbrace{\alpha \boldsymbol Z^*_{\ell}\boldsymbol z_{\ell}-\left(\boldsymbol I + \alpha \boldsymbol{Z}_{\ell}^* \boldsymbol{Z}_{\ell}\right)\left[\boldsymbol{q}_{\ell}\right]_{\star}}_{=0}\right) = \boldsymbol z_{\ell}\\
\end{align}
This proves that $\boldsymbol E_{\ell}^{-1}\boldsymbol y_{\ell} = \boldsymbol z_{\ell}$. The statement that you are looking for follows immediately.
