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Is the following problem convex function?

\begin{eqnarray} \begin{aligned} \underset{\mathbf{X}} {\text{minimize}}\,\,\,\, & \text{Trace}(\mathbf{RX}) \\ \text{subject to} &\\ & \text{Trace}(\mathbf{Q_{k}X}) = {0}\,\, k = 1,2...N\\ &\mathbf{AX} = \mathbf{0} \\ & \text{Trace}(\mathbf{X}) = {1} \\ & X\succeq 0,\,\, X \in \mathbb{R}^{M \times M}. \end{aligned} \end{eqnarray} Where, $\mathbf{R} \in \mathbb{C}^{M \times M}$ is a positive definite symmetric matrix.

and

$\mathbf{Q_{k}} \in \mathbb{R}^{M \times M}\,\, k = 1,2,...N$ are the symmetric Toeplitz matrices, like the matrices given below \begin{eqnarray} \begin{aligned} Q_1 &= \left[\begin{array}{cccccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right]\\ \end{aligned} \end{eqnarray} and \begin{eqnarray} \begin{aligned} Q_2 &=\left[\begin{array}{cccccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right] \end{aligned} \end{eqnarray} In General, $ Q_k$ = Toep( [0 0 0... 1 ... 0\, 0] ), Where $1$ is at $k^{th}$ position.

and

$\mathbf{A} \in \mathbb{R}^{V \times M}$, $V < \frac{M}{2}$.

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1 Answer 1

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It's convex.

Note that $trace(A X)$ is just a linear function. So you're minimising a linear function subject to a bunch of linear equality constraints and a positive semidefiniteness constraint.

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