Prove that the spectral radius $\rho(A)$ is a continuous function, where $A$ is a square matrix.

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.

• (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). Dec 5, 2018 at 14:15
• @freakish Thanks. But how to prove part(b)? Dec 5, 2018 at 14:17
• 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. Dec 5, 2018 at 19:14
• (specifically the arzela-ascoli theorem) Dec 5, 2018 at 19:17
• I think you can find a solution in this forum. I remember seeing a proof using complex analysis. It seems that Rouche theorem applies. Dec 5, 2018 at 22:38