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Let $||. ||$ be some norm on $\mathbb{R^n}.$ Interpret the real-valued matrices as a Euclidean space, $\mathbb{R^{n^2}}=Mat_{n\times n},$ and prove that the following are continuous functions of the $n^2$ matrix entries $A_{ij}$

(a) The spectral radius of $A$, $\rho(A)=\max\{|\lambda| ~~| \text{such that}~\lambda ~\text{is an eigenvalue of}~A \}.$

(b) The spectral radius of the inverse, $f(A)=\rho (A^{-1}),$ on the domain of invertible matrices.


For part (b), I think the goal is $\displaystyle\lim_{\Delta A \to 0} \rho(A+\Delta A)=\rho(A).$ I tried to use Gelfand formula. Then I have $\rho(A)=\displaystyle\lim_{\Delta A \to 0} \rho(A+\Delta A)=\displaystyle\lim_{\Delta A \to 0} \lim_{k\to \infty}||(A+\Delta A)^k||^{1/k},$ which becomes more complicated. In addition, I tried to use the operator norm to bound the spectral radius, but I felt that I was still far away from the proof.

Why does the part (b) need to prove if the part (a) is true? $f(A)=\rho(A) $ restrict on the domain of invertable matrices. I was confuled about part(b).


I know how to prove this statement using a theorem in complex analysis: The roots of a complex-valued polynomial are continuous wrt the coefficients of the polynomial.

By using this theorem, (b) is very easy to prove.

I am looking for another proof of (b) without the theorem in complex analysis.

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  • $\begingroup$ (b) follows from (a) because $A\mapsto A^{-1}$ is continuous on $GL_n(\mathbb{R})$ regardless of norm (all of them are equivalent anyway). $\endgroup$
    – freakish
    Dec 5, 2018 at 14:15
  • $\begingroup$ @freakish Thanks. But how to prove part(b)? $\endgroup$
    – learner
    Dec 5, 2018 at 14:17
  • $\begingroup$ Does anyone know whether the functions $f_k:M_{n\times n}\rightarrow \mathbb{R}$, $f_k(A) = \|A^k\|^{\frac{1}{k}}$, have derivatives $\frac{\partial f_k}{\partial a_{ij}}$ uniformly bounded (in $k$) when $f_k$'s are restricted to some compact set of matrices? Then one could use limit theorems to show spectral radius is continuous. $\endgroup$
    – Matthew C
    Dec 5, 2018 at 19:14
  • $\begingroup$ (specifically the arzela-ascoli theorem) $\endgroup$
    – Matthew C
    Dec 5, 2018 at 19:17
  • $\begingroup$ I think you can find a solution in this forum. I remember seeing a proof using complex analysis. It seems that Rouche theorem applies. $\endgroup$ Dec 5, 2018 at 22:38

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