considering following quadratic optimization problem with linear equality constraints \begin{equation} \begin{aligned} \max_w~& r^\top w - \frac{1}{2}w^\top Q w \\ \mathrm{s.t.}& \sum w = 1 \end{aligned} \end{equation} By given $r\in\Re^{N\times 1}$ and positive definite $Q$, it is easy to find the analytical solution: \begin{equation} \begin{bmatrix} -Q & \mathbf{1}_{N\times 1} \\ \mathbf{1}_{1\times N} & 0 \end{bmatrix}\begin{bmatrix} w\\ \lambda \end{bmatrix} = \begin{bmatrix} r\\ 1 \end{bmatrix} \end{equation} where $\lambda$ is the Lagrange multiplier. I was wondering, by given the optimal $w^*$, is it possible to derive the vector $r$?
I thought it would be easy, because \begin{equation} w = Q^{-1} (r + \mathbf{1}_{N\times 1} \lambda) \Longrightarrow \sum(w) = \mathbf{1}_{1\times N}Q^{-1} (r + \mathbf{1}_{N\times 1} \lambda) = 1 \end{equation} hence \begin{equation} \begin{bmatrix} I_N & \mathbf{1}_{N\times 1} \\ \mathbf{1}_{1\times N}Q^{-1} & \mathbf{1}_{1\times N}Q^{-1}\mathbf{1}_{N\times 1} \end{bmatrix}\begin{bmatrix} r\\ \lambda \end{bmatrix} = \begin{bmatrix} Qw\\ 1 \end{bmatrix} \end{equation} However, I found that matrix is not invertible due to rank deficiency. Can anyone help me to obtain the $r$ given $w^*$? Thanks!
The problem you pose is minimization of $||w-Q^{-1/2}r||_{Q^{1/2}}$ over the simplex, i.e. some weighted projection to the simplex. Given a projection $w^{*}$ there are generically infinitely many $r$ which yield the same projection (thus the issue with non-invertible). A picture in 2d says it all hopefully (here drawn for simple case with $W=I$, red line is the simplex, green line all possible $r$, two centers $r^{*}$ with corresponding level sets of the quadratic, and both of them yield the same projection $w^{*}$