I'm struggling to show the following statement:let $K$ be a field and $\lambda_i,i=1,..,r\in K$ eigenvalues of matrix $A \in K^{n\times n}$ and $m_i,j=1,..,r$ it's algebraic multiplicities. Let $m_1+..+m_r=n$. I need to show that $\text{Tr} (A)=\sum_{i=1}^rm_i\lambda_i$.
We observe that $\det(A-xI_n)=(\lambda_1-x)^{m_1}\cdot ...\cdot (\lambda_r-x)^{m_r}$. We know as well that the coefficient of power $n-1$ in the characteristic polynomial of $A$ is the trace of $A$. I see why the statement is correct, but I don't really see how to justify it in the correct way. If someone could help, I would really appreciate it. Thank you!
Jordan Form
At least for $\mathbb C$-valued matrices the way to go about this is to prove the relation for the Jordan form (obvious) and then use the relation $Tr(MN) = Tr(NM)$ on the Jordan decomposition. Meaning $A = S^{-1} J S = MN$ for $M = S^{-1} J$ and $N = S$ and so $$Tr(A)\!= \!Tr(S^{-1} J S) \!= \!Tr(MN) \!=\! Tr(NM) \!= \!Tr (SS^{-1} J) = Tr(J) = \sum_{i=1}^r m_i \lambda_i.$$
Whether this works for general fields depends on how well the Jordan form generalises to fields that might not be algebraically closed.