Inverse of diagonal dominant matrices

[geoffrey159]

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

Homework Equations

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 ?

 

[wabbit]

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

 

[geoffrey159]

Thank you