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Math Help - Max norm

  1. #1
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    Max norm

    I have no idea how to tackle this problem.Let A be an n \times n matrix. The matrix 2-norm and max norm are defined as \|A\|_2=max \frac{\|Ax\|_2}{\|x\|_2} and \|A\|_{max}=max \mid a_{i,j} \mid
    Show that \|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 n \times n matrix. The matrix 2-norm and max norm are defined as \|A\|_2=max \frac{\|Ax\|_2}{\|x\|_2} and \|A\|_{max}=max \mid a_{i,j} \mid
    Show that \|A\|_{max} \leq \|A\|_2 \leq n\|A\|_{max}
    let A=[a_{ij}]. fix 1 \leq j \leq n and suppose e_j, is the n \times 1 vector with 1 in the j-th entry and 0 anywhere else. then ||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 x=\begin{bmatrix}x_1 & \cdots & x_n \end{bmatrix}^T \neq 0. first note that by Cauchy-Schwarz we have: \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:

    \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 x \neq 0, we'll get ||A||_2 \leq n||A||_{\max}. \ \Box
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