With the definition of $ \lVert A \rVert_2$ and $\lVert A \rVert_1$ and $\lVert A \rVert_ \infty$ that is:

\begin{gather} \lVert A\rVert_1 = \max_{j} \sum_{i=1}^m \lvert a_{ij}\rvert\\ \lVert A\rVert_2 = \sqrt{\rho(A^HA)}\\ \lVert A\rVert_\infty = \max_{i} \sum_{j=1}^n \lvert a_{ij}\rvert \end{gather}

prove that:

$$\lVert A \rVert_2^2 \leq \lVert A \rVert_1 \lVert A \rVert_\infty$$

  • 1
    $\begingroup$ Rather than \parallel, use \lVert and \rVert for the norms, that gives proper spacing when rendered. $\endgroup$ – Daniel Fischer Feb 18 '15 at 13:11

Let $\|\cdot\|$ be any matrix norm induced by a vector norm. Then we have $$\|A\|_2^2= \rho(AA^H) \leq \|AA^H\| \leq \|A\|\|A^H\|.$$ Here the first inequality follows from a "famous theorem" (see e.g. Proposition 4.4) and the second inequality follows from the fact that $\|\cdot\|$ is a matrix norm induced by a vector norm and thus is submultiplicative. Finally note that $\|A\|_1 =\|A^H\|_\infty$.


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