Proving that $det(A) ne 0$ with $A$ satisfying following conditions.











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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.










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  • It looks like a Laplacian matrix with positive weight. But Laplacian matrix $mathbf{L}$ satisfies $sum_{i} L_{ij} = sum_{j} L_{ij} = 0$ which gives $det(mathbf{L}) = 0$.
    – K_inverse
    2 days ago












  • @dmsj djsl If $det (A) =0$ then the columns of $A $ are linearly dependent, as suggested by the hint. I'd try to use this i n conjunction with the three conditions to get a contradiction.
    – AnyAD
    2 days ago








  • 1




    Google "diagonally dominant matrix".
    – darij grinberg
    2 days ago















up vote
4
down vote

favorite
1












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.










share|cite|improve this question






















  • It looks like a Laplacian matrix with positive weight. But Laplacian matrix $mathbf{L}$ satisfies $sum_{i} L_{ij} = sum_{j} L_{ij} = 0$ which gives $det(mathbf{L}) = 0$.
    – K_inverse
    2 days ago












  • @dmsj djsl If $det (A) =0$ then the columns of $A $ are linearly dependent, as suggested by the hint. I'd try to use this i n conjunction with the three conditions to get a contradiction.
    – AnyAD
    2 days ago








  • 1




    Google "diagonally dominant matrix".
    – darij grinberg
    2 days ago













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





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.










share|cite|improve this question













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.







linear-algebra abstract-algebra matrices determinant






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asked 2 days ago









dmsj djsl

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32317












  • It looks like a Laplacian matrix with positive weight. But Laplacian matrix $mathbf{L}$ satisfies $sum_{i} L_{ij} = sum_{j} L_{ij} = 0$ which gives $det(mathbf{L}) = 0$.
    – K_inverse
    2 days ago












  • @dmsj djsl If $det (A) =0$ then the columns of $A $ are linearly dependent, as suggested by the hint. I'd try to use this i n conjunction with the three conditions to get a contradiction.
    – AnyAD
    2 days ago








  • 1




    Google "diagonally dominant matrix".
    – darij grinberg
    2 days ago


















  • It looks like a Laplacian matrix with positive weight. But Laplacian matrix $mathbf{L}$ satisfies $sum_{i} L_{ij} = sum_{j} L_{ij} = 0$ which gives $det(mathbf{L}) = 0$.
    – K_inverse
    2 days ago












  • @dmsj djsl If $det (A) =0$ then the columns of $A $ are linearly dependent, as suggested by the hint. I'd try to use this i n conjunction with the three conditions to get a contradiction.
    – AnyAD
    2 days ago








  • 1




    Google "diagonally dominant matrix".
    – darij grinberg
    2 days ago
















It looks like a Laplacian matrix with positive weight. But Laplacian matrix $mathbf{L}$ satisfies $sum_{i} L_{ij} = sum_{j} L_{ij} = 0$ which gives $det(mathbf{L}) = 0$.
– K_inverse
2 days ago






It looks like a Laplacian matrix with positive weight. But Laplacian matrix $mathbf{L}$ satisfies $sum_{i} L_{ij} = sum_{j} L_{ij} = 0$ which gives $det(mathbf{L}) = 0$.
– K_inverse
2 days ago














@dmsj djsl If $det (A) =0$ then the columns of $A $ are linearly dependent, as suggested by the hint. I'd try to use this i n conjunction with the three conditions to get a contradiction.
– AnyAD
2 days ago






@dmsj djsl If $det (A) =0$ then the columns of $A $ are linearly dependent, as suggested by the hint. I'd try to use this i n conjunction with the three conditions to get a contradiction.
– AnyAD
2 days ago






1




1




Google "diagonally dominant matrix".
– darij grinberg
2 days ago




Google "diagonally dominant matrix".
– darij grinberg
2 days ago










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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).






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    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).






    share|cite|improve this answer

























      up vote
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      down vote



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      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).






      share|cite|improve this answer























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        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).






        share|cite|improve this answer












        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).







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        answered 2 days ago









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