Is there an explicit relationship between the eigenvalues of a matrix and its derivative? If we consider a matrix $A$ dependent of a variable $x$, the eigenvalues and eigenvectors satisfying the equation
$$
A \vec{v}=\lambda \vec{v}
$$
will also depend on $x$. If we consider the matrix $B$ such that 
$$B_{ij}=\frac{ \mathrm{d}}{ \mathrm{d} x} A_{ij}$$ 
Then, could we express the eigenvalues of $B$ in terms of the eigenvalues of $A$?
I found the question very interesting and was not able to find a satisfying answer myself.
For example in the case for $2\times2$ matrices of the form
$$
A=\left (
\begin{matrix}
 a(x) & b(x) \\ 
 0 & c(x) 
\end{matrix}
\right ),\implies B=\left (
\begin{matrix}
 a'(x) & b'(x) \\ 
 0 & c'(x) 
\end{matrix}
\right )
$$
I noticed that $\lambda_B(x)= \lambda_A'(x)$. But I cannot generalise it to general $2\times 2$ matrices. Not even thinking about $n\times n$ matrices...
Thank you for your help and any idea!
 A: It is not true in general that the eigenvalues of $B(x)$ are the derivatives of those of $A(x)$. And this even for some square matrices of dimension $2 \times 2$.
Consider the matrix
$$A(x) =
\begin{pmatrix}
1& -x^2\\
-x &1
\end{pmatrix}$$
It’s characteristic polynomial is $\chi_{A(x)}(t)=t^2-2t+1-x^3$, which has for roots $1\pm x ^{3/2}$ for $x>0$. Those are the eigenvalues of $A(x)$.
The derivative of $A(x)$ is 
$$B(x) =
\begin{pmatrix}
0& -2x\\
-1 &0 
\end{pmatrix}$$ and it’s characteristic polynomial is $\chi_{B(x)}(t)=t^2-2x$, whose roots are $\pm \sqrt{2} x^{1/2}$for $x>0$.
We get a counterexample as the derivative of $1+x^{3/2}$ is not $\sqrt{2}x^{1/2}$.
However in the special case of upper triangular matrices (that you consider in your original question) the eigenvalues of the matrix derivative are indeed the derivatives of the eigenvalues.
A: It can be shown that the eigenvalues of the derivative of the matrix cannot be derived from the eigenvalues of the original matrix. Example:
$$
A_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
\;\;\; , \;\;\; 
A_2 = \begin{pmatrix} 0 & e^x \\ e^{-x} & 0 \end{pmatrix}
$$
Both of the matrices above have the eigenvalues $-1$ and $1.$ However, the derivative of the first matrix has the eigenvalue $0$ (with multiplicity 2), while the derivative of the second matrix has the eigenvalues $i$ and $-i.$ 
Just given the eigenvalues $-1$ and $1$, there is no way of telling which matrix they originate from, hence no way of getting the eigenvalues of the derivative.
A: Let $\{\alpha_k,\beta_k\}$ be the eigenvalues of $(A,B)$
where $B(x) = A'(x).$
A class of matrices for which $\beta_k=\alpha_k'\,$ can be constructed as follows.
Choose an orthogonal matrix $Q$ and an upper triangular matrix $U(x).$
$$\eqalign{
A &= Q\,U\,Q^{-1} \cr
A' &= Q\,U'Q^{-1} \cr
}$$
Since $Q$ is orthogonal, the eigenvalues of $(U,U')$ equal the eigenvalues of $(A,A')$, respectively.
Since the eigenvalues of a triangular matrix are its diagonal elements,
the EVs of $U$ are $\{U_{kk}=\alpha_k\}\,$ and
 the EVs of $U'$ are $\{U'_{kk}=\alpha_k'=\beta_k\}.$
NB: $\,Q$ must be independent of $x$ for this construction to apply.
A: If $A(x)v= \lambda(x)v$, with v independent of x, then, differentiating on both sides of the equation by x, $A'(x)v= \lambda'(x)v$.  That is, the eigenvalues of A' are the derivatives of the eigenvalues of A, with same associated eigenvectors.
