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Inverse of diagonal dominant matrices

  1. Jun 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that ##n\times n ## complex matrices such that ##\forall 1\le i \le n,\quad \sum_{k\neq i} |a_{ik}| < |a_{ii}|##, are invertible

    2. Relevant equations


    3. The attempt at a solution

    If I show that the column vectors are linearly independent, then the matrix has rank ##n## and is invertible.

    Let ##\lambda_1, ..., \lambda_n## be complex coefficients such that ##\sum_{k = 1}^n \lambda_k a_{ik} = 0## for all ##1 \le i \le n##. There must be ##\lambda_i## such that ## |\lambda_j| \le |\lambda_i| ## for all ##j\neq i##. If ##\lambda_i \neq 0##, then ## |a_{ii} | = \frac{1}{|\lambda_i|} | \sum_{k\neq i} \lambda_k a_{ik} | \le \frac{|\lambda_i|}{|\lambda_i|} \sum_{k\neq i} |a_{ik}| < |a_{ii}|##
    This is absurd so ##\lambda_i = 0## and all the other lambda's are zero because ##\lambda_i## dominates in module all the other lambda's. So the column vectors are linearly independant.

    Is it Ok ?
     
  2. jcsd
  3. Jun 8, 2015 #2

    wabbit

    User Avatar
    Gold Member

    Yes, your proof is concise and to the point - nothing to add.
     
  4. Jun 8, 2015 #3
    Thank you
     
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