Let $A$ be a $n \times n$ real matrix, $(\lambda_i, v_i)$ be the $i$-th (eigenvalue, eigenvector) of $A^T$, and $x(t)$ be a vector of $n$ functions $x_i(t)$. For $\frac{d x(t)}{dt}=A x(t)$, the following holds
$$ \frac{d (\prod_i v_i^T x(t))}{dt} = \sum_i \left( \frac{d(v_i^T x(t))}{dt} ~ \prod_{j \neq i} v_j^T x(t)\right) = \sum_i \left( v_i^T A x(t) ~ \prod_{j \neq i} v_j^T x(t)\right) = \sum_i \left( (A^T v_i)^T x(t) ~ \prod_{j \neq i} v_j^T x(t)\right) = \sum_i \left( (\lambda_i v_i)^T x(t) ~ \prod_{j \neq i} v_j^T x(t)\right) = \sum_i \left( \lambda_i v_i^T x(t) ~ \prod_{j \neq i} v_j^T x(t)\right) = \sum_i \left( \lambda_i \prod_j v_j^T x(t)\right) = \left(\sum_i \lambda_i\right) \left(\prod_i v_i^T x(t)\right) = tr(A) \prod_i v_i^T x(t)$$
Thus, for $f(t) = \prod_i v_i^T x(t)$ we have $\frac{d f(t)}{dt} = tr(A) f(t)$. We do not need to know the particular eigenvalues of $A$ in order to express the derivative of $f(t)$. In other words, we can express the derivative of $f(t)$ using the elements of $A$.
Is there a way to rewrite $f(t)$ such that only elements of $A$ are used, e.g. based on elementary symmetric polynomials?
Remark: $\sum_i$ is a shortcut for $\sum_{i=1}^n$, and $\prod_i$ is a shortcut for $\prod_{i=1}^n$.
Certainly, we can solve the differential equation $$ \DeclareMathOperator{\tr}{trace} f'=\tr(A)f $$ To conclude that $$ f(t)=C\exp[\tr(A)t] $$ For some constant $C$, which depends on the matrix $A$. In particular, $$ C= \prod_i v_i^Tx(0) $$ I can't think of a way to write that in terms of the matrix entries, though.