Let $A_1,A_2$ be real matrices with negative eigenvalues. The the Lyapunov operators
$$L_1(P) = A_1^T\cdot P + P\cdot A_1$$ and
$$ L_2(P) = A_2^T\cdot P + P\cdot A_2 $$ are nonsingular. Let
$$ P_1 = \int_{t=0}^{\infty} e^{t\cdot A_1^T}\cdot Q_1 \cdot e^{t\cdot A_1} \cdot dt$$ be the solution to the equation $L_1(P) = -Q_1$ with $Q_1 \succeq 0$. Then consider
$$ Q_2 = e^{-t\cdot A_2^T} \cdot e^{t\cdot A_1^T} \cdot Q_1 \cdot e^{t\cdot A_1}\cdot e^{-t\cdot A_2} \succeq 0$$ and
$$P_2 = \int_{t=0}^{\infty} e^{t\cdot A_2^T}\cdot Q_2 \cdot e^{t\cdot A_2} = P_1$$ is the solution to $L_2(P) = -Q_2 \preceq 0$. Then consider the matrix:
$$ A = \alpha \cdot A_1 +(1-\alpha)\cdot A_2$$ with $\alpha \in (0,1)$. We obtain
$$ A^T\cdot P + P\cdot A = \alpha (A_1^T\cdot P + P\cdot A_1) + (1-\alpha)(A_2^T\cdot P + P\cdot A_2) \\
= -\alpha\cdot Q_1 - (1-\alpha)\cdot Q_2 \preceq 0$$
It seems like a proof that a convex combination of two stable matrices is also stable. However, I read, this is not true in general, so my question is where is my mistake?
Remark
$Q_2$ is not constant ...