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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!

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    $\begingroup$ One nice property of an upper (or lower) triangular matrix is that its eigenvalues are the same as its diagonal elements; which explains the $2\times 2$ examples that you discovered. But this is not a general property of all matrices. $\endgroup$ – greg Jan 27 at 20:51
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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.

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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.

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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.

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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.

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    $\begingroup$ The bias with what you say is that in general, the eigenvector depends on $x$. $\endgroup$ – mathcounterexamples.net Jan 25 at 20:49

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