# Inverse of diagonal dominant matrices

## Homework Statement

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

## 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 ?

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wabbit
Gold Member
Yes, your proof is concise and to the point - nothing to add.

• geoffrey159
Thank you