Show that if \(A=(a_{ij})\) is an \(n\times n\) complex matrix satisfying \[ |a_{ii}|>\sum_{j\ne i}|a_{ij}| \] for all \(i\in\{1,\dots,n\}\), then \(A\) is invertible.
Solution
Assume \(A x = 0\) and choose i such that \(\left| x _ { i } \right| = \operatorname* { m a x } _ { j } \left| x _ { j } \right|\) . Then
\[ | a _ { i i } | | x _ { i } | \leq \sum _ { j \neq i } | a _ { i j } | | x _ { j } | \leq \sum _ { j \neq i } | a _ { i j } | | x _ { i } | \]
so that
\[ \left( \left| a _ { i i } \right| - \sum _ { j \neq i } \left| a _ { i j } \right| \right) \left| x _ { i } \right| \leq 0 . \]
Since the first factor is positive by assumption and the second is nonnegative, we must have \(x _ { i } = 0\) . By choice of i we must have \(x = 0\) so A is invertible.