Let $L/K$ be a finite separable extension of degree $n := [L:K]$. Define $m_x : L \to L$ as the $K$-linear transformation $m_x(a) = ax$, where $a, x \in L$. We define the trace of $x$ as $$\operatorname{Tr}_{L/K}(x) := \operatorname{Tr}(m_x)$$ whereby the trace on the right hand side is the trace of the matrix representing $m_x$.
Until now I've been able to show that $\operatorname{Tr}_{K(x)/K}(x) = \sum_{i=1}^{[K(x):K]} \sigma_i(x)$ where the $\sigma_i \in \operatorname{Hom}_K(K(x), \bar{K})$, so for a primitive element $x$, and I've also been able to show (I think) that $$\operatorname{Tr}_{L/K}(x) = [L:K(x)]\sum_{i=1}^r\sigma_i(x)$$, where $r = [K(x):K]$, by considering a block diagonal companion matrix with each of the blocks equal to the companion matrix for the minimal polynomial of $x$ over $K$.
My question is: I don't know how to go from this step to concluding that
$$\operatorname{Tr}_{L/K}(x) = \sum_{i=1}^{[L:K]}\sigma_i(x).$$
If I simply "require" that $L = K(x)$ then that solves the issue with the $[L:K(x)]$ coefficient, but how do I know that there are $[L:K(x)]$ of the $\sigma_i$ in this case?
Since $L/K$ is separable, $L/K(x)$ is separable, hence any $K$-algebra morphism $\tau:K(x)\to K_{alg}$ extends in exactly $[L:K(x)]$ $K$-algebra morphisms $L:\to K_{alg}$ (property of finite separable extensions).
Since $x\in K(x)$, you have exactly $[L:K(x)]$ $K$-algebra morphisms $\sigma:L\to K_{alg}$ such that $\sigma(x)=\sigma_i(x)$, and you are done.