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Let $A(z)$ be a $n\times n$ square matrix depending on the complex value $z$ and $\lambda_z$ is its spectral radius.

Is $\lambda_z$ continous or is it possible that it can jump? Or maybe someone knows a good example for that?

Comment: $A(z)$ consists of complex functions $a_{ij}(z)$ and the spectral radius is defined as $$ \lambda_z=\max\{|\lambda|:\text{ $\lambda$ is an eigenvalue of $A(z)$}\}. $$

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    $\begingroup$ How do you define the largest eigenvalue $\lambda\in\mathbb C$? $\endgroup$
    – Mikhail Katz
    Commented Sep 6, 2023 at 11:43
  • $\begingroup$ Sorry, I updated the answers to your questions! $\endgroup$
    – Fynn13
    Commented Sep 6, 2023 at 11:46
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    $\begingroup$ certainly not; for a simple counter example, take the $2\times 2 $ diagonal matrix with $iz$ and $z^2$ on the diagonal; as $z$ moves along the real axis and crosses 1 the "largest" eigenvalue jumps from $i$ to 1. $\endgroup$
    – Carlo Beenakker
    Commented Sep 6, 2023 at 11:55
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    $\begingroup$ a counter example with real matrix elements: $2\times 2$ diagonal matrix with $-x$ and $x^2$ on the diagonal, the eigenvalue with the largest absolute value jumps from $-1$ to $1$ when $x$ crosses 1. $\endgroup$
    – Carlo Beenakker
    Commented Sep 6, 2023 at 12:22
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    $\begingroup$ the eigenvalues themselves depend continuously on the matrix elements, so if all eigenvalues are positive taking the absolute value makes no difference and the largest eigenvalue will depend continuously on the matrix elements. $\endgroup$
    – Carlo Beenakker
    Commented Sep 6, 2023 at 12:34

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$\newcommand\la\lambda$If (all the entries $a_{ij}(z)$ of) the matrix $A(z)$ are continuous in $z$, then, yes, $\la_z$ is continuous in $z$.

Indeed, take any complex $z$ and any sequence $(z_k)$ converging to $z$. Then, according to (say) Sections 5.1 and 5.2 of Chapter II, the set of eigenvalues of $A(z_k)$ converges to the set of eigenvalues of $A(z)$ in the following sense: for some enumeration $(\la_1(z),\dots,\la_n(z))$ of the eigenvalues of $A(z)$ and some enumerations $(\la_1(z_k),\dots,\la_1(z_k))$ of the eigenvalues of $A(z_k)$ one has $$\max(|\la_1(z_k)-\la_1(z)|,\dots,|\la_n(z_k)-\la_n(z)|)\to0$$ (as $k\to\infty$). It follows that $$\la_{z_k}=\max(|\la_1(z_k)|,\dots,|\la_n(z_k)|) \to\max(|\la_1(z)|,\dots,|\la_n(z)|)=\la_z.\quad\Box$$

Clearly, the condition that $A(z)$ be continuous in $z$ cannot be dropped. For example, consider the case when $a_{11}(z)$ is not continuous in $z$ whereas $a_{ij}(z)=0$ for all $z$ if $(i,j)\ne(1,1)$.

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  • $\begingroup$ By Gelfand's formula, the spectral radius can be computed as the limit of operator norms of the powers of the matrix, but I don't see immediately how to derive continuity. There should be a simpler argument than enumerating all the eigenvalues. $\endgroup$
    – Mikhail Katz
    Commented Sep 6, 2023 at 14:37
  • $\begingroup$ @MikhailKatz : (i) I think the argument is already very simple. (ii) To deal with the maximum of the moduli of all the eigenvalues, we need to deal with all the eigenvalues somehow. Of course, enumerating them is only one way to do that. Alternatively, one can e.g. describe the continuity of the set of eigenvalues the way it is done in Section 5.1 of Chapter II of Kato's book, linked in my answer. (iii) Like you, I don't see a simple (or any other) way to use Gelfand's formula here. $\endgroup$
    – Iosif Pinelis
    Commented Sep 6, 2023 at 15:14
  • $\begingroup$ @MikhailKatz: Using just Gelfand's formula can't do the job since Gelfand's formula also holds in infinite dimensions while contuity of the spectral radius does not. $\endgroup$
    – Jochen Glueck
    Commented Sep 6, 2023 at 17:02
  • $\begingroup$ @JochenGlueck : This is a good point! $\endgroup$
    – Iosif Pinelis
    Commented Sep 6, 2023 at 17:43

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