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!

  • 2
    $\begingroup$ 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, 2019 at 20:51

4 Answers 4


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.


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.


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.


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.

  • 5
    $\begingroup$ The bias with what you say is that in general, the eigenvector depends on $x$. $\endgroup$ Jan 25, 2019 at 20:49

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