# Optimizing over vector and Matrix at the same time.

I want to know if my understand (prove convexity, format as SDP) the following problem is correct: \begin{equation*} \begin{aligned} & \min_{c\in \mathbb{C}^n,D\in \mathbb{H}_+^n} && \|c\|_2 + \lambda \cdot \mathrm{Tr}(D) \\ & \text{subject to} && h^Tc + \mathrm{Tr}(H^TD) = b\\ &&& D \succeq 0 \end{aligned} \end{equation*} where $$D \in \mathbb{H}_+^n$$ is a Hermitian Positve-Semidefinite Matrix.

This problem is derived from Trace Heuristic of Matrix rank-minimization, but is regularized with a vector $$c$$ and the equality constraint has term $$c$$ in it.

## Proof Convexity

To show that the problem above is a minimizing a convex function over a convex set, I show that the objective is jointly convex in $$(c,D)$$ and the equality constraint admits a convex set over $$(c,D)$$. The SDP cone constraint is convex and does not need to be shown.

From here, I write $$x = (c,D) \in \mathbb{C}^n \times \mathbb{H}_+^n$$ and write $$\mathbf{X} = \mathbb{C}^n \times \mathbb{H}_+^n$$

### Convex Objective

For any $$\theta \in [0,1]$$, for given $$x_1=(c_1,D_1),x_2=(c_2,D_1) \in \mathbf{X}$$, we denote convex combination as $$x_3 = \theta x_1 + (1-\theta) x_2$$, noting that $$x_3 =(c_3,D_3) \in \mathbf{X}$$.

$$$$\begin{split} f(\theta x_1 + (1-\theta) x_2) &= \|\theta c_1 + (1-\theta)c_2\|_2 +\lambda \cdot \mathrm{Tr}(\theta D_1 + (1-\theta) D_2) \\ &\leq \theta\|c_1\|_2 +(1-\theta)\|c_2\|_2 + \theta \lambda \mathrm{Tr}(D) + (1-\theta)\lambda\mathrm{Tr}(D) \\ &= \theta f(x_1) + (1+\theta)f(x_2) \end{split}$$$$

### Convex Set

Simply by the affine nature of the equality.

## SDP

Let $$\mathbf{Y} = \begin{bmatrix} diag(c) & 0 \\ 0 & D\end{bmatrix}$$ Re-formulate the problem as : \begin{equation*} \begin{aligned} & \min_{Y\in \mathbb{H}_+^{2n}} && \langle \begin{bmatrix} diag(c) & 0 \\ 0 & \lambda I_n\end{bmatrix} , Y\rangle_{\mathbb{H}^n_+} \\ & \text{subject to} && \langle \begin{bmatrix} diag(h) & 0 \\ 0 & H\end{bmatrix}, Y\rangle_{\mathbb{H}^n_+} =b \\ &&& Y \succeq 0 \quad (\dagger) \end{aligned} \end{equation*} ($$\dagger$$): This inequality is stronger in the sense that it forces $$Re(c) \geq 0$$ whereas the original formulation has no such constraint.

## My Questions

1. Is my proof of convexity correct?
2. How to formulate the problem as an SDP? Specifically, what do with the inequality constraint?
• do you mean tr($H^TD$) in the constraint of the initial problem? Jan 30, 2019 at 19:55
• @LinAlg Yep! Just corrected, thanks! Jan 30, 2019 at 20:36

2. You need to write $$c$$ as the difference of two nonnegative vectors (resulting in an extra block row/column in $$Y$$). This is the same as formulating a linear optimization problem in standard form where it initially has a free variable.
• 1. When you say a convex set defined by convex function, do you mean that for $h_i(x) = 0$, $h_i$ has to be convex in $x$? So in my case I will need to argue that $h^Tc + Tr(H^T D)$ is convex in $(c,D)$? Revisiting Boyd 4.2.1 seem to suggest that equality constraint needs to be linear in most cases, unless the problem satisfies specific conditions as in Exercise 4.6. As per objective, I've just always had trouble when thinking with multiple variables. sum of convex function is convex always look to me like $f(x) + g(x)$ over the same $x$. Jan 30, 2019 at 20:52
• equality constraints indeed need to be linear, for an inequality constraint $g_i(x)\leq 0$, $g$ needs to be convex; these statements are stronger than "[the feasible region is a] Convex Set". Jan 30, 2019 at 20:53
• Thanks! As per the second part, re-writing $c = x^{+}-x^{-}, Re(x^+) \succeq 0, Re(x^-) \succeq 0$ and defining $Y = \begin{bmatrix} diag(x^+) & 0 & 0 \\ 0 & diag(x^-) & 0 \\ 0 & 0& D\end{bmatrix}$ does the trick! For proof of convexity, I'm afraid I don't understand what need to be added to the my proof. Would the statement that the equality constraint is linear suffice? For inequality $g(x) \leq 0$, is the convexity required necessary because some functions can admit convex set under $g(x) \leq 0$ but not under $g(x) \leq \alpha$ (quasiconvex)? Could you give an pathological example? Jan 30, 2019 at 21:08
• "Would the statement that the equality constraint is linear suffice?" yes. "Could you give an pathological example?" it can be quite simple: $\{x : x^3 \leq 0\}$ is a convex set, but $g(x)=x^3$ is not a convex function Jan 30, 2019 at 21:19