Jtdylktuy

I am given $A in M_n(mathbb{R})$ which satisfies the following conditions.

1. 
   $A_{i,i} gt 0$ for all $1 le i le n$
   


2. 
   $A_{i,j} le 0$ for all distinct $1 le i, j le n$
   


3. 
   $sum_{j=1}^n A_{i,j} gt 0$ for all $1 le i le n$
   

Then, I am supposed to show that $det(A) ne 0$

Now, I am frankly not sure where to even start. However, I was given the following hint:

If not, there is a non-zero solution of $Ax = 0$. If $x_i$ has largest absolute value, show that the $i$th linear equation from $Ax=0$ leads to a contradiction.

I don't really quite get how to apply this hint either. Could someone help? Thanks.

Let $Ax = 0$. Then, let $x_i = arg max |x_j|$ i.e. $i$ is such that $|x_i| geq |x_j|$ for all $i neq j$. By assumption, if $x neq 0$ then $|x_i| > 0$.

Note that $Ax = 0$ implies that $A_i cdot x = 0$, where $A_i$ denotes the $i$th row of $A$, as a vector. This follows from the definition of matrix multiplication.

However, $A_i cdot x = sum_{j} A_{ij}x_j$. By definition, we have $|x_i| geq |x_j|$ for all $j$, so write $$A_i cdot x = A_{ii}x_i + sum_{j neq i} A_{ij}x_j$$ and use the inequality $|x+y| geq |x| - |y|$, to see that :
$$
|A_i cdot x| geq |A_{ii}x_i| - left|sum_{j neq i} A_{ij}x_jright|
$$

But, we know that $|x_j| leq |x_i|$, so it follows that $$|sum_{j neq i} A_{ij}x_j| leq sum_{j neq i} -A_{ij}|x_j| leq -|x_i|sum_{j neq i}A_{ij}$$.

Therefore,
$$
|x_i|A_{ii} - left|sum_{j neq i} A_{ij}x_jright| geq |x_i| times sum_{j} A_{ij} > 0
$$

Which is a contradiction, since $A_i cdot x = 0$. Consequently, no such $x$ exists.

More can be said. Indeed, the Gerschgorin circle theorem guarantees that every eigenvalue lies with a Gerschgorin disc, whose centre is one of the diagonal entries, and radius is the sum of the absolute values of the non-diagonal entries of the row. In this case, by the conditions given, the theorem gives that no eigenvalue can in fact be smaller than the smallest value of $sum_{j} A_{ij}$, which is greater than $0$. So this way the result is clear.

Also, the matrix with conditions given is strictly diagonally dominant, and from the Gerschgorin circle theorem is non-singular (known as the Levy-Desplanques theorem, and having applications in probability).