This is easy to formulate as a semidefinite programming problem.

First, let $X=xx^{T}$. The semidefiniteness constraint becomes

$A-\lambda X \succeq 0$

Next, use a standard technique to handle the absolute value in the objective by replacing it with an auxiliary variable and two linear inequality constraints. The problem becomes

$\min_{\lambda,t} t $

subject to

$t \geq \lambda-\lambda_{0} $

$t \geq \lambda_{0}-\lambda $

$A-\lambda X \succeq 0$

If $t$ is greater than or equal to $\lambda-\lambda_{0}$ and $t$ is greater than or equal to $\lambda_{0}-\lambda$, then $t$ is clearly greater than or equalt to $| \lambda-\lambda_{0} |$. Since $t$ is being minimized and there are no other constraints on $t$, it will end up equal to $| \lambda-\lambda_{0}|$.

This isn't quite in standard SDP format. The two constraints involving $t$ can be brought into semidefinite form by making

$t - \lambda + \lambda_{0} $

and

$t - \lambda_{0} + \lambda $

diagonal elements of the matrix that is constrained to be positive semidefinite. This insures that $t-\lambda+\lambda_{0} \geq 0$ and $t-\lambda_{0}+\lambda \geq 0$.

Let

$ F_{0}=\left[ \begin{array}{ccc} A & 0 & 0 \\\ 0 & \lambda_{0} & 0 \\\ 0 & 0 & -\lambda_{0} \end{array} \right] $

$ F_{1}=\left[ \begin{array}{ccc} -X & 0 & 0 \\\ 0 & -1 & 0 \\\ 0 & 0 & 1 \end{array} \right] $

$F_{2}=\left[ \begin{array}{ccc} 0 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{array} \right] $

Now, the problem can be written in standard form as

$\min_{\lambda,t} t $

subject to

$F_{0}+\lambda F_{1}+tF_{2} \succeq 0$

This is easy to formulate as a semidefinite programming problem.

First, let $X=xx^{T}$. The semidefiniteness constraint becomes

$A-\lambda X \succeq 0$

Next, use a standard technique to handle the absolute value in the objective by replacing it with an auxiliary variable and two linear inequality constraints. The problem becomes

$\min_{\lambda,t} t $

subject to

$t \geq \lambda-\lambda_{0} $

$t \geq \lambda_{0}-\lambda $

$A-\lambda X \succeq 0$

If $t$ is greater than or equal to $\lambda-\lambda_{0}$ and $t$ is greater than or equal to $\lambda_{0}-\lambda$, then $t$ is clearly greater than or equal to $| \lambda-\lambda_{0} |$. Since $t$ is being minimized and there are no other constraints on $t$, it will end up equal to $| \lambda-\lambda_{0}|$.

This isn't quite in standard SDP format. The two constraints involving $t$ can be brought into semidefinite form by making

$t - \lambda + \lambda_{0} $

and

$t - \lambda_{0} + \lambda $

diagonal elements of the matrix that is constrained to be positive semidefinite. This insures that $t-\lambda+\lambda_{0} \geq 0$ and $t-\lambda_{0}+\lambda \geq 0$.

Let

$ F_{0}=\left[ \begin{array}{ccc} A & 0 & 0 \\\ 0 & \lambda_{0} & 0 \\\ 0 & 0 & -\lambda_{0} \end{array} \right] $

$ F_{1}=\left[ \begin{array}{ccc} -X & 0 & 0 \\\ 0 & -1 & 0 \\\ 0 & 0 & 1 \end{array} \right] $

$F_{2}=\left[ \begin{array}{ccc} 0 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{array} \right] $

Now, the problem can be written in standard form as

$\min_{\lambda,t} t $

subject to

$F_{0}+\lambda F_{1}+tF_{2} \succeq 0$

This is easy to formulate as a semidefinite programming problem.

First, let $X=xx^{T}$ (it's a constant so easy to compute.) The semidefiniteness constraint becomes

$A-\lambda X \succeq 0$

Next, use a standard technique to handle the absolute value in the objective by replacing it with an auxiliary variable and two linear inequality constraints. The problem becomes

$\min_{\lambda,t} t $

subject to

$t \geq \lambda-\lambda_{0} $

$t \geq \lambda_{0}-\lambda $

$A-\lambda X \succeq 0$

If $t$ is greater than or equal to $\lambda-\lambda_{0}$ and $t$ is greater than or equal to $\lambda_{0}-\lambda$, then $t$ is clearly greater than or equalt to $| \lambda-\lambda_{0} |$. Since $t$ is being minimized and there are no other constraints on $t$, it will end up equal to $| \lambda-\lambda_{0}|$.

This isn't quite in standard SDP format. The two constraints involving $t$ can be brought into semidefinite form by making

$t - \lambda + \lambda_{0} $

and

$t - \lambda_{0} + \lambda $

diagonal elements of the matrix that is constrained to be positive semidefinite. This insures that $t-\lambda+\lambda_{0} \geq 0$ and $t-\lambda_{0}+\lambda \geq 0$.

Let

$ F_{0}=\left[ \begin{array}{ccc} A & 0 & 0 \\\ 0 & \lambda_{0} & 0 \\\ 0 & 0 & -\lambda_{0} \end{array} \right] $

$ F_{1}=\left[ \begin{array}{ccc} -X & 0 & 0 \\\ 0 & -1 & 0 \\\ 0 & 0 & 1 \end{array} \right] $

$F_{2}=\left[ \begin{array}{ccc} 0 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{array} \right] $

Now, the problem can be written in standard form as

$\min_{\lambda,t} t $

subject to

$F_{0}+\lambda F_{1}+tF_{2} \succeq 0$

This is easy to formulate as a semidefinite programming problem.

First, let $X=xx^{T}$. The semidefiniteness constraint becomes

$A-\lambda X \succeq 0$

Next, use a standard technique to handle the absolute value in the objective by replacing it with an auxiliary variable and two linear inequality constraints. The problem becomes

$\min_{\lambda,t} t $

subject to

$t \geq \lambda-\lambda_{0} $

$t \geq \lambda_{0}-\lambda $

$A-\lambda X \succeq 0$

If $t$ is greater than or equal to $\lambda-\lambda_{0}$ and $t$ is greater than or equal to $\lambda_{0}-\lambda$, then $t$ is clearly greater than or equalt to $| \lambda-\lambda_{0} |$. Since $t$ is being minimized and there are no other constraints on $t$, it will end up equal to $| \lambda-\lambda_{0}|$.

This isn't quite in standard SDP format. The two constraints involving $t$ can be brought into semidefinite form by making

$t - \lambda + \lambda_{0} $

and

$t - \lambda_{0} + \lambda $

diagonal elements of the matrix that is constrained to be positive semidefinite. This insures that $t-\lambda+\lambda_{0} \geq 0$ and $t-\lambda_{0}+\lambda \geq 0$.

Let

$ F_{0}=\left[ \begin{array}{ccc} A & 0 & 0 \\\ 0 & \lambda_{0} & 0 \\\ 0 & 0 & -\lambda_{0} \end{array} \right] $

$ F_{1}=\left[ \begin{array}{ccc} -X & 0 & 0 \\\ 0 & -1 & 0 \\\ 0 & 0 & 1 \end{array} \right] $

$F_{2}=\left[ \begin{array}{ccc} 0 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{array} \right] $

Now, the problem can be written in standard form as

$\min_{\lambda,t} t $

subject to

$F_{0}+\lambda F_{1}+tF_{2} \succeq 0$