[Theorem I, Taussky]

Lemma 55.2.1. Let $A = (a_{ij})$ be a complex $n \times n$ matrix.

1. If $|a_{ii}| > \sum _{j \not= i} |a_{ij}|$ for each $i$, then $\det (A)$ is nonzero.

2. If there exists a real vector $m = (m_1, \ldots , m_ n)$ with $m_ i > 0$ such that $|a_{ii} m_ i| > \sum _{j \not= i} |a_{ij}m_ j|$ for each $i$, then $\det (A)$ is nonzero.

Proof. If $A$ is as in (1) and $\det (A) = 0$, then there is a nonzero vector $z$ with $Az = 0$. Choose $r$ with $|z_ r|$ maximal. Then

$|a_{rr} z_ r| = |\sum \nolimits _{k \not= r} a_{rk}z_ k| \leq \sum \nolimits _{k \not= r} |a_{rk}||z_ k| \leq |z_ r| \sum \nolimits _{k \not= r} |a_{rk}| < |a_{rr}||z_ r|$

which is a contradiction. To prove (2) apply (1) to the matrix $(a_{ij}m_ j)$ whose determinant is $m_1 \ldots m_ n \det (A)$. $\square$

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