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

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.

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.

• I am new to this subject. Could you explain what $inf({||A^n||^{1 / n} : n \in N})$ means? Mar 19, 2019 at 8:50
• $\inf$ means infimum.
– Fred
Mar 19, 2019 at 8:51
• To prove this, can we also say: let $\lambda$ be the largest (in magnitude) eigenvalue of a square matrix $B$, and let $v$ be corresponding eigenvector with $||v||_1=1.$ Then $$|λ|=||λv||_1=||Bv||_1≤||B||_1||v||_1=||B||_1$$ Dec 11, 2020 at 14:39

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

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

• Sorry, I made a mistake in the question, I have updated it. Mar 19, 2019 at 8:20