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Thread: Max norm

  1. #1
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    Max norm

    I have no idea how to tackle this problem.Let A be an $\displaystyle n \times n$ matrix. The matrix 2-norm and max norm are defined as $\displaystyle \|A\|_2=max \frac{\|Ax\|_2}{\|x\|_2}$ and $\displaystyle \|A\|_{max}=max \mid a_{i,j} \mid$
    Show that $\displaystyle \|A\|_{max} \leq \|A\|_2 \leq n\|A\|_{max}$
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    I have no idea how to tackle this problem.Let A be an $\displaystyle n \times n$ matrix. The matrix 2-norm and max norm are defined as $\displaystyle \|A\|_2=max \frac{\|Ax\|_2}{\|x\|_2}$ and $\displaystyle \|A\|_{max}=max \mid a_{i,j} \mid$
    Show that $\displaystyle \|A\|_{max} \leq \|A\|_2 \leq n\|A\|_{max}$
    let $\displaystyle A=[a_{ij}].$ fix $\displaystyle 1 \leq j \leq n$ and suppose $\displaystyle e_j,$ is the $\displaystyle n \times 1$ vector with 1 in the j-th entry and 0 anywhere else. then $\displaystyle ||A||_2 \geq \frac{||Ae_j||_2}{||e_j||_2}=\sqrt{|a_{1j}|^2 + \cdots + |a_{nj}|^2} \geq |a_{ij}|, \ \ \forall \ 1 \leq i \leq n.$

    this proves the left hand side of the inequality. for the right hand side, let $\displaystyle x=\begin{bmatrix}x_1 & \cdots & x_n \end{bmatrix}^T \neq 0.$ first note that by Cauchy-Schwarz we have: $\displaystyle \frac{\left|\sum_{j=1}^n a_{ij}x_j \right|^2}{\sum_{j=1}^n |x_j|^2} \leq \sum_{j=1}^n|a_{ij}|^2.$ thus:

    $\displaystyle \frac{||Ax||_2}{||x||_2} \leq \sqrt{\sum_{i=1}^n \sum_{j=1}^n |a_{ij}|^2} \leq \sqrt{n^2(\max |a_{ij}|)^2}=n \max |a_{ij}|=n ||A||_{\max}.$ since the inequality holds for any $\displaystyle x \neq 0,$ we'll get $\displaystyle ||A||_2 \leq n||A||_{\max}. \ \Box$
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