Spectral Radius $\leq$ min(1-norm, infinity norm)

Question

How do I prove that the spectral radius of a matrix is less than or equal to the minimum of 1-norm and infinity norm of the matrix? i.e. $$\rho(A) \leq min(||A||_1, ||A||_{\infty})$$

I know the inequalities between matrix norms i.e. $$ ||A||_{\infty} \leq ||A||_p \leq n^{\frac{1}{p}} ||A||_{\infty}$$ $$ ||A||_{2} \leq ||A||_1 \leq \sqrt{n} ||A||_{2}$$ But, I am not sure how this will help me prove the above equation.

Answer 1

If $|| \cdot||$ is a submultiplicative matrix-norm, then we have

$$(*) \quad\rho(A)= \lim_{n \to \infty}||A^n||^{1/n}= \inf \{||A^n||^{1/n}: n \in \mathbb N\} \le ||A||.$$

The norms $|| \cdot||_1$ and $|| \cdot||_{\infty}$ are both submultiplicative matrix-norms, hence, by $(*)$:

$\rho(A) \le || A||_1$ and $\rho(A) \le || A||_{\infty}$. This gives the result.

Answer 2

$\rho(A) = \min(||A||_1, ||A||_{\infty})$ is not true !

Take a nilpotent matrix $A \ne 0$. Then $\rho(A)=0$, but $\min(||A||_1, ||A||_{\infty})>0.$