questions about Gerschgorin circle theorem. Q:if $A$ is a strictly diagonally dominant matrix, prove that
$$|\det A|\ge \prod_{i=1}^n(|a_{ii}|-\sum_{j\neq i}|a_{ij}|)$$
the proof is:
by  Gerschgorin circle theorem, the eigenvalue of $A$ lies in the union of the following circles:
$$|z-a_{ii}|\le\sum_{j\neq i}|a_{ij}|\quad i=1,\cdots,n$$
so the eigenvalue $\lambda$ satisfy $$|\lambda|\ge|a_{ii}|-\sum_{j\neq i}|a_{ij}|$$
and we get the conclusion.

my question is that the last equation only holds for some $i$, for example, all the eigenvalue can lie in the same circle, so the conclusion does not hold. What's the problem?
 A: Take the entries of the matrix to be $A = (a_{ij})_{n \times n}$
Note that the system of linear equaltions:
$$a_{i1} + \sum\limits_{j=2}^{n} a_{ij}x_j= 0,\qquad i = 2,3,\cdots, n$$
has unique solution, say $(x_2,x_3,\cdots, x_n)$ (by the given condition on $A$ it follows that the determinant of the system is non singular, by Girschgorin Theorem).
So, $$\begin{align}-a_{ii}x_i = a_{i1}+\sum\limits_{j=2,j \neq i}^{n} a_{ij}x_j &\implies |a_{ii}| \le \frac{|a_{i1}|}{|x_i|} +\sum\limits_{j=2,j \neq i}^{n} |a_{ij}|.\frac{|x_j|}{|x_i|}\\&\implies |a_{i1}| +\sum\limits_{j=2,j \neq i}^{n} |a_{ij}| < \frac{|a_{i1}|}{|x_i|} +\sum\limits_{j=2,j \neq i}^{n} |a_{ij}|.\frac{|x_j|}{|x_i|}\\&\implies \max_{2 \le i\le n} |x_i| < 1\end{align}$$
Now, consider the determinant $\det \Delta_1$ of the system and add the $j^{th}$ column multiplied by $x_j$ to the first column for each $j = 2,3,\cdots, n$.
Thus, the determinant $\det \Delta_1$ becomes:
$$\det \Delta_1 = \left(a_{11}+\sum\limits_{j=2}^{n} a_{1j}x_j\right).\det \left(\begin{matrix} a_{22} & \cdots & a_{2n}\\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot  \\ a_{n2} & \cdots & a_{nn}\end{matrix}\right) = \left(a_{11}+\sum\limits_{j=2}^{n} a_{1j}x_j\right) \times \det \Delta_2$$
Since, $|x_j| < 1$ for each $j = 2(1)n$, we have:
$$|\det \Delta_1| \ge \left(|a_{11}|-\sum\limits_{j=2}^{n} |a_{1j}|\right).|\det \Delta_2|$$
Again $\det \Delta_2$ is also diagonally dominant, so the conclusion follows by induction.
