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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.

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2 Answers 2

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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.

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  • $\begingroup$ I am new to this subject. Could you explain what $inf({||A^n||^{1 / n} : n \in N})$ means? $\endgroup$
    – akshayk07
    Mar 19, 2019 at 8:50
  • $\begingroup$ $\inf$ means infimum. $\endgroup$
    – Fred
    Mar 19, 2019 at 8:51
  • $\begingroup$ 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 $$|λ|=||λv||_1=||Bv||_1≤||B||_1||v||_1=||B||_1$$ $\endgroup$
    – user56202
    Dec 11, 2020 at 14:39
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$\rho(A) = \min(||A||_1, ||A||_{\infty})$ is not true !

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

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  • $\begingroup$ Sorry, I made a mistake in the question, I have updated it. $\endgroup$
    – akshayk07
    Mar 19, 2019 at 8:20

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