Solve $\min_{\mathbf{x},\mathbf{R}}\left\|\mathbf{y}-\mathbf{x}\right\|_{\mathbf{R}}^2$ s.t. $\mathbf{A}\mathbf{x}=\mathbf{0}$; $\mathbf{R}\succeq0$ Dear optimization experts, 
Do you have a suggestion how to solve this problem?
\begin{align}
\min_{\mathbf{x}, \mathbf{R}, \mathbf{\lambda}}  f(\mathbf{x}, \mathbf{R}) 
&= \left\| \mathbf{y} - \mathbf{x} \right\|_{\mathbf{R}}^2 + \mathbf{\lambda}^T \mathbf{A}\mathbf{x}  \\
&= \left( \mathbf{y} - \mathbf{x}  \right)^T \mathbf{R} \left( \mathbf{y} - \mathbf{x}  \right) + \mathbf{\lambda}^T \mathbf{A}\mathbf{x},
\end{align} 
where $\mathbf{y} \in \mathbb{R}^{n \times 1}$ is given, $\mathbf{x} \in \mathbb{R}^{n \times 1}$, $\mathbf{R} \in \mathbb{R}^{n \times n}$, and $\mathbf{\lambda} \in \mathbb{R}^{n \times 1}$ is a Lagrange Multiplier, and $\mathbf{A} \in \mathbb{R}^{n \times n}$ is given. 
If it is dependent on a single variable (and Lagrange multiplier), that is either $\mathbf{x}$ or $\mathbf{R}$, then it can be solved in closed-form. But I am not sure how to do it for more than one variable jointly, in this case. Thank you very much in advance.
EDIT:
Original constrained optimization problem
\begin{align*}
\text{minimize}_{\mathbf{x}, \mathbf{R}} \quad \ & \left( \mathbf{y} - \mathbf{x}  \right)^T \mathbf{R} \left( \mathbf{y} - \mathbf{x}  \right) \\
\text{subject to }\quad & \mathbf{A} \mathbf{x} = \mathbf{0} 
\end{align*}
EDIT EDIT:
$\mathbf{R} \succeq 0$ is positive semi-definite (constraint).

If $\mathbf{R}$ is fixed and invertible (so it should be positive definite to be on safe side), then I can show that (also suggested by Amrit P.)
\begin{align}
\mathbf{x} = \mathbf{y} - \left(\mathbf{R} + \mathbf{R}^T \right)^{-1} \mathbf{A}^T \mathbf{\lambda} \ ,
\end{align}
and $\mathbf{\lambda}$ can be obtained by plugging $\mathbf{x}$ to the constraint 
\begin{align}
\mathbf{A}\mathbf{x} = \mathbf{0} \ ,
\end{align}
such that
\begin{align}
\mathbf{\lambda} = \left(\mathbf{A} \left(\mathbf{R} + \mathbf{R}^T \right)^{-1}  \mathbf{A}^T \right)^{-1} \mathbf{A} \mathbf{x} \ .
\end{align}

I am still thinking, what if we want to find optimal $\mathbf{R}$ as well... perhaps I am hitting the wall :( ... 
I saw a paper that also obtain $R$ but not jointly .... e.g., cf. R. Kumar and A. Tyagi, "Weighted Least Squares Based Spectral Precoder for OFDM Cognitive Radio", in IEEE Wireless Comm. Letters, Dec. 2015.
 A: Are you sure that $R$ is a variable in this optimization? Without that condition, it's a relatively straightforward problem.
Since you have already made the constraint a part of the optimization objective, we can take partial derivatives wrt variables of the problem ($x,R,\lambda$), and set them to zero for obtaining the optimal point.
$$\frac{\partial{f(x,R,\lambda)}}{\partial x}=-(y-x)^T(R+R^T)+\lambda^TA=0\tag{1}$$
$$\frac{\partial{f(x,R,\lambda)}}{\partial R}=(y-x)(y-x)^T=0\tag{2}$$
$$\frac{\partial{f(x,R,\lambda)}}{\partial\lambda}=x^TA=0\tag{3}$$
Equation $2$ suggests that $x=y$, which makes sense since that would immediately set the positive part of the objective to $0$.
Equation $3$ suggests that $x$ should lie in the null-space of $A^T$. Combining equations $2$ and $3$ tells us that an optimal solution exists only if $y$ belongs to the null space of $A^T$:
$$y=\sum_ic_ia_i^{\perp}=A^{\perp}c$$
where $a_i^{\perp}$ is the column vector of $A^{\perp}:A^TA^{\perp}=0$.
Now let's consider the case where the above is not true, i.e.
$$y=A^{\perp}c+v:v=Ad\neq0$$
In such a case, if you were to run the above through a convex optimizer, you would end up with-
$$x=y+\epsilon:A(v+\epsilon)=0$$
$$\min_{\epsilon,R,\lambda}\epsilon^TR\epsilon+\lambda^T(Av+A\epsilon)$$
$$\implies 2R\epsilon+A^T\lambda=0\Leftrightarrow\epsilon=-\frac{1}{2}R^{-1}A^T\lambda$$
$$A\epsilon=-Av\Leftrightarrow AR^{-1}A^T\lambda=2Av$$
We can calculate $\lambda$ and $\epsilon$ uniquely if $AR^{-1}A^T$ is invertible.
