Let $A_1, A_2 \in GL_n(\mathbb R)$ be two fixed matrices with eigenvalues lying on the open left half plane of $\mathbb C$, i.e., with negative real parts and $A_1 \neq A_2$. We may assume $A_1, A_2$ are diagonalizable if necessary. Let $(A_1)_\alpha$ and $(A_2)_\beta$ be the corresponding matrix that scales the eigenvalues by $\alpha \in \mathbb R$ and $\beta \in \mathbb R$ respectively. That is if $(\lambda_1, \dots, \lambda_n)$ are eigenvalues of $A_1$, then $(\alpha \lambda_1, \dots, \alpha \lambda_n)$ are eigenvalues of $(A_1)_{\alpha}$.
I am wondering whether it is possible to construct a continuous path $\gamma: [0,1] \to GL_n(\mathbb R)$ with eigenvalues all lying on the left half plane along the path: by choosing a sequence of nonzero real numbers, $(\alpha_n)$ and $(\beta_n)$, $\gamma$ is piecewise linear. For example, suppose there exist some finite sequences $(\alpha_1, \dots, \alpha_k)$ and $(\beta_1, \dots, \beta_l)$, then the path would be \begin{align*} A_1 \xrightarrow{} (A_1)_{\alpha_1} \xrightarrow{} \dots \xrightarrow{} (A_1)_{\alpha_k} \xrightarrow{} (A_2)_{\beta_1} \xrightarrow{} \dots \xrightarrow{} (A_2)_{\beta_l} \xrightarrow{} A_2. \end{align*} In between, each arrow stands for a convex path. Assume we can construct a continuous path between $A$ and $(A)_{\alpha}$.
If we take a convex combination of $A_1, A_2$, i.e., $A(t) = (1-t)A_1 + tA_2$. Then $\det(A(t))$ is a nonconstant polynomial in $t$ and has at most $n$ zeros. Intuitively I am thinking of taking the path until we meet a singular point of $t$ (by continuity, all eigenvalues should stay in the left half plane) and change path to some $(A_1)_{\alpha}$ and doing this repeatedly. Not sure whether this will work.
p.s. The reason for my consideration was because I need to guarantee certain matrix structure along the path. I have worked out scaling eigenvalues can be done within the structure of interest.