Proof of ADMM special case

Question

I was reading the textbook about ADMM and came across some special cases without proof, which is really interesting so I was trying to prove them. I did a couple but had some trouble getting this one so please help. Thank you in advance

Considering x-update when A=I

$x^+=argmin_x(f(x)+(ρ/2)∣∣x−v∣∣_2^2)=prox_{f,ρ}(v)$.

The special case to be proved is $f(x)=(1/2)x^TPx+q^Tx+r, x^+:= (P+{\rho}A^TA)^{-1}({\rho}A^Tv-q)$

The attached photo shows what I have tried so far, I think I am close but there is something not right.

Answer 1

Hint (almost full solution)

Assuming you want to solve the following update

\begin{align} x^{+} = \arg\min_{x} \frac{1}{2} x^TPx + q^Tx + r + \frac{\rho}{2}\| Ax - b - v \|_2^2. \end{align}

Compute the gradient and set it to zero to obtain the $x$-update solution.

• Gradient of $\frac{1}{2} x^TPx + q^Tx+ r$ is $\frac{1}{2}(P+P^T)x + q$ (see example here How to take the gradient of the quadratic form?). Assuming $P$ is symmetric, the gradient simplifies to $Px + q$.
• Gradient of the last piece, i.e., $\frac{\rho}{2}\|Ax-b-v\|^2$, is $\rho A^T(Ax-b-v)$.

To this end, add the gradients and set them to zero such that \begin{align} Px + q + \rho A^T(Ax-b-v) = 0 \end{align}

Can you take it from here?