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Let $A$ be a strictly diagonally dominant matrix of dimensions $n \times n$. ("Strictly diagonally dominant" means that $\left|a_{i,i}\right| > \sum\limits_{j \neq i} \left|a_{i,j}\right|$ for all $i \in \left\{1,2,\ldots,n\right\}$, where $a_{u,v}$ denotes the $\left(u,v\right)$-th entry of $A$.)

Prove that $A$ is invertible.

My attempt builds on the proof of Gershgorin's circle theorem, given in the Wikipedia article https://en.wikipedia.org/wiki/Gershgorin_circle_theorem

Let $\lambda$ be an eigenvalue of $A$, and scale its corresponding eigenvector $x$ so that $x_{i} = 1$ and $|x_{j}| \leq 1$ for $j \neq i$

Then $Ax = \lambda x$, and in particular $\sum_{i\neq j}a_{i,j}x_{j} = \lambda - a_{ii}$ (1)

Now because $A$ is strictly diagonally dominant it holds for every $i$ that,

$\sum_{j \neq i}|a_{i,j}| < |a_{i,i}|$ and since $|x_{i}| \leq 1$ the following should hold:

$\sum_{i\neq j}|a_{i,j}x_{j}| \leq \sum_{j \neq i}|a_{i,j}| < |a_{i,i}|$

But here I get stuck, feel like I want to use (1) in some way to complete the proof and put $\lambda = 0$ to get a contradiction, thus proving that if $A$ is strictly diagonally dominant, it has non-zero eigenvalues which should imply invertibility.. Am I in the right direction?

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    $\begingroup$ Note that you have proved that $$|\lambda-a_{i,i}|\leq \sum_{j\not =i}|a_{i,j}|<|a_{i,i}|$$ $\endgroup$ – Kelenner Sep 8 '17 at 12:30
  • $\begingroup$ The approach is fine (not sure about the details). The mentioned theorem and the triangle inequality should lead to the desired proof. $\endgroup$ – Peter Sep 8 '17 at 13:48
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For an elementary proof, assume there exists a vector $x \ne 0$ such that $Ax = 0$. This implies $\sum_{j=1}^n a_{ij}x_j = 0, \forall i \in \{1, \ldots, n\}$. Let $x_k = \|x\|_\infty \ne 0$, i.e. $x_k$ is the the largest entry of $x$ by absolute value.

We have:

$$0 = \sum_{j=1}^n a_{kj}x_j \implies a_{kk}x_k = -\sum_{j\ne k} a_{kj}x_j \implies a_{kk} = -\sum_{j\ne k} a_{kj}\frac{x_j}{x_k}$$

By taking the absolute value we get:

$$|a_{kk}| = \left|\sum_{j\ne k} a_{kj}\frac{x_j}{x_k}\right| \leq \sum_{j\ne k} |a_{kj}|\underbrace{\left|\frac{x_j}{x_k}\right|}_{\leq 1} \leq \sum_{j\ne k} |a_{kj}|$$

This is a contradiction since $A$ is strictly diagonally dominant.

This means that $0\notin \sigma(A)$, hence $A$ is invertible.

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