I am trying to understand how to solve the following simple semidefinite program by hand.
$$\begin{array}{ll} \text{minimize} & x_{11}\\ \text{subject to} & 2 x_{12} = 1\\ & \begin{pmatrix} x_{11} & x_{12}\\ x_{12} & x_{22} \end{pmatrix} \succeq 0\end{array}$$
I have figured out that in the standard formulation we have
$$\begin{array}{ll} \text{minimize} & \mbox{tr} \left( C X \right)\\ \text{subject to} & \mbox{tr} \left( A X \right) = b\\ & X \succeq 0\end{array}$$
where
$$ C = \left(\begin{array}{cc}1 & 0\\ 0 & 0 \end{array}\right)$$ and $$ A = \left(\begin{array}{cc}0 &1\\ 1 & 0 \end{array}\right)$$ and $ b = 1$
However, I have no idea how to proceed from there. Any pointers?
Because of the restriction we get $ X = \left( \begin{array}{cc} x_{11} & 1/2 \\ 1/2 & x_{22} \end{array} \right)$ and therefore $x_{11}x_{22} - 1/4 \geq 0 $ because $X$ is positive semidefinite. This is the solution for the primal form. However I have no idea how to calculate the solution for the dual form.