The question is a bit unclear here, so I'll start by restating the problem. The OP has a matrix
$A=\left[
\begin{array}{cc}
0.5 & -0.4 \\
1 & -0.5
\end{array}
\right]$
and wants to find a symmetric matrix $Q$ such that
$Q-A^{T}QA \succeq 0$
$Q \succ 0$
The problem here is that $Q \succ 0$ isn't an SDP constraint. We could put in $Q \succeq 0$, but $Q=0$ is a solution to
$Q-A^{T}QA \succeq 0$
$Q \succeq 0$
and we want to constrain $Q$ to not be singular. In the comments, the OP suggested solving the SDP
$\min t$
$Q-A^{T}QA \succeq 0$
$tI-Q-A^{T}QA \succeq 0$
$Q \succeq 0$.
but $t=0$, $Q=0$ is an optimal solution to this problem too!
An alternative SDP formulation that does ensure that $Q \succ 0$ is
$\min 0$
$Q-A^{T}QA - \epsilon I \succeq 0$
$Q \succeq 0$
where $\epsilon$ is a small positive constant (I'll use $\epsilon=0.01$ in the following.) Because $A^{T}QA$ is positive semidefinite and $\epsilon I$ is positive definite, these constraints enforce $Q \succ 0$ (In fact, $Q \succeq \epsilon I$, so the smallest eigenvalue of $Q$ will be greater than or equal to $\epsilon$.)
Modeling packages like CVX and Yalmip can easily turn this formulation into an SDP that can be solved by a variety of solvers such as SDPT3, SeDuMi, CSDP, SDPA, etc. However, the OP wants to see how to turn this into a standard form SDP
$\max \mbox{tr}(CX) $
$\mbox{tr}(A_{i}X)=b_{i}\;\; i=1, 2, \ldots, m$
$X \succeq 0$
The first step is to introduce a slack variable $S$, and write the
problem as
$\max 0$
$Q-A^{T}QA-S=\epsilon I$
$S \succeq 0$
$Q \succeq 0$.
The constraint $Q-A^{T}QA-S=\epsilon I$ is linear in the elements of $Q$, although this might not be immediately obvious. The key observation is that
$A^{T}QA=\sum_{i=1}^{2} \sum_{j=1}^{2} Q_{i,j} A_{i,:}^{T}A_{j,:}$
Since the matrices in $Q-A^{T}QA-S$ are all symmetric, we need 3 linear equality constraints for the (1,1), (1,2) and (2,2) elements of the matrix equality.
$0.75Q_{1,1}-0.5Q_{1,2}-0.5Q_{2,1}-1.0Q_{2,2}-S_{1,1}=\epsilon$
$0.2Q_{1,1}+1.25Q_{1,2}+0.4Q_{2,1}+0.5Q_{2,2}-S_{1,2}=0$
$-0.16Q_{1,1}-0.2Q_{1,2}-0.2Q_{2,1}+0.75Q_{2,2}-S_{2,2}=\epsilon$
The constraints must be in symmetric form, so we rewrite the second constraint as
$0.2Q_{1,1}+0.825Q_{1,2}+0.825Q_{2,1}+0.5Q_{2,2}-0.5S_{1,2}-0.5S_{2,1}=0$
Finally, we embed $Q$ and $S$ into a block diagonal matrix
$X=\left[
\begin{array}{cc}
Q & 0 \\
0 & S \\
\end{array}
\right].$
The problem becomes
$\max \mbox{tr}(CX)$
$\mbox{tr}(A_{1}X)=\epsilon$
$\mbox{tr}(A_{2}X)=0$
$\mbox{tr}(A_{3}X)=\epsilon$
$X \succeq 0$
where
$C=0$
$A_{1}=\left[
\begin{array}{cccc}
0.75 & -0.5 & 0 & 0 \\
-0.5 & -1.0 & 0 & 0 \\
0 & 0 & -1.0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right]$
$A_{2}=\left[
\begin{array}{cccc}
0.2 & 0.825 & 0 & 0 \\
0.825 & 0.5 & 0 & 0 \\
0 & 0 & 0 & -0.5 \\
0 & 0 & -0.5 & 0 \\
\end{array}
\right]$
$A_{3}=\left[
\begin{array}{cccc}
-0.16 & -0.2 & 0 & 0 \\
-0.2 & +0.75 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1.0 \\
\end{array}
\right]$
In SDPA format, the problem is
3
2
2 2
0.01 0 0.01
1 1 1 1 0.75
1 1 1 2 -0.5
1 1 2 2 -1.0
1 2 1 1 -1.0
2 1 1 1 0.2
2 1 1 2 0.825
2 1 2 2 0.5
2 2 1 2 -0.5
3 1 1 1 -0.16
3 1 1 2 -0.2
3 1 2 2 +0.75
3 2 2 2 -1.0
I solved this problem using CSDP and obtained
$Q=\left[
\begin{array}{cc}
38.5170 & -11.8166 \\
-11.8166 & 24.3371 \\
\end{array}
\right]$
$S=\left[
\begin{array}{cc}
16.3572 & 0.3746 \\
0.3746 & 16.8067 \\
\end{array}
\right]$
It's easy to verify that $Q$ has all of the required properties.
There are infinitely many solutions to this problem- there is no reason to expect that all solvers will return the same solution.
You could also adjust the objective function or add additional constraints to push the solution in some desired direction. For example, you might want to minimize the maximum eigenvalue of $Q$ or minimize the sum of the eigenvalues of $Q$.