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We are given the problem \begin{equation}\label{primal} \begin{aligned} & \underset{x}{\text{minimize}} & & x^TA_0x+2b_0^Tx+c_0 \\ & \text{subject to} & & x^TA_1x+2b_1^Tx+c_1 \leq 0 \\ \end{aligned} \end{equation}
where the $A_i$ are symmetric matrices which are NOT asssumed to be positive semi-definite; this is why this is a non-convex optimisation problem.

I am able to derive its dual problem using Schur complement trick as \begin{equation}\label{dual} \begin{aligned} & \underset{\gamma, \lambda}{\text{maximize}} & & \gamma \\ & \text{subject to} & & \lambda \geq 0 \\ & & & \begin{bmatrix}A_0+\lambda A_1 & b_0 + \lambda b_1 \\ (b_0 + \lambda b_1)^T & c_0 + \lambda c_1 - \gamma\end{bmatrix} \succeq 0 \end{aligned} \end{equation} Now the reference claims that the dual problem of the SDP is an SDP with variables $X\in S^n, x\in R^n$ \begin{equation}\label{dualdual} \begin{aligned} & \underset{x, X}{\text{minimize}} & & {tr}(A_0X)+2b_0^T x + c_0\\ & \text{subject to} & & {tr}(A_1X)+2b_1^Tx+c_1 \leq 0\\ & & & \begin{bmatrix}X & x \\ x^T & 1\end{bmatrix} \succeq 0 \end{aligned} \end{equation}
How can I derive this?

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Taking hints form this Derive the dual of the semidefinite program $\max_Y{\rm Tr}(X^T Y)$ subject to $\begin{pmatrix}I&Y\\ Y^T&I\end{pmatrix}\succeq0$

I was able to work out the following: We introduce Lagrange variables $\alpha$ and $\begin{bmatrix} X & x \\ x^T & \beta\end{bmatrix}$ and write the Lagrangian as $\mathcal L(\gamma, \lambda, \alpha, X, x, \beta) = -\gamma - \langle \alpha, \lambda \rangle -\langle\begin{bmatrix} X & x \\ x^T & \beta\end{bmatrix}, \begin{bmatrix}A_0+\lambda A_1 & b_0 + \lambda b_1 \\ (b_0 + \lambda b_1)^T & c_0 + \lambda c_1 - \gamma\end{bmatrix}\rangle$. Now the dual function is $g(\alpha, X, x, \beta)=\inf_{\gamma, \lambda}\mathcal L(\gamma, \lambda, \alpha, X, x, \beta) = \begin{cases}-\langle X, A_0\rangle-2\langle x, b_0\rangle-\langle\beta, c_0\rangle & -I + \beta =0, \ \alpha + \langle X, A_1\rangle+2\langle x, b_1\rangle+\langle\beta, c_1\rangle \\ -\infty & otherwise\end{cases}$. Now, substituting $\beta=I$ and using $\alpha\succeq 0$ this can be written as the following dual problem

\begin{equation} \begin{aligned} & \underset{x, X}{\text{minimize}} & & {tr}(A_0X)+2b_0^T x + c_0\\ & \text{subject to} & & {tr}(A_1X)+2b_1^Tx+c_1 \leq 0\\ & & & \begin{bmatrix}X & x \\ x^T & 1\end{bmatrix} \succeq 0 \end{aligned} \end{equation}

The last inequality is the condition that Lagrangian multiplier should be PSD.

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