Inverse of strictly diagonally dominant matrix
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I have a matrix whose diagonal entries are positive whereas non-diagonal entries are negative.This matrix is also Strictly diagonally dominant.
Can we say that all elements of the inverse of this matrix is strictly positive i.e $a_{ij}$>0 .
linear-algebra matrices
asked Dec 30 '18 at 4:08
Ayush MishraAyush Mishra
485
$endgroup$
add a comment |
$begingroup$
I have a matrix whose diagonal entries are positive whereas non-diagonal entries are negative.This matrix is also Strictly diagonally dominant.
Can we say that all elements of the inverse of this matrix is strictly positive i.e $a_{ij}$>0 .
linear-algebra matrices
asked Dec 30 '18 at 4:08
Ayush MishraAyush Mishra
485
$endgroup$
$begingroup$
See en.wikipedia.org/wiki/M-matrix where it is said in the 3 first lines that it is a known result (but without proof)
$endgroup$
– Jean Marie
Dec 30 '18 at 18:41
add a comment |
$begingroup$
I have a matrix whose diagonal entries are positive whereas non-diagonal entries are negative.This matrix is also Strictly diagonally dominant.
Can we say that all elements of the inverse of this matrix is strictly positive i.e $a_{ij}$>0 .
linear-algebra matrices
asked Dec 30 '18 at 4:08
Ayush MishraAyush Mishra
485
$endgroup$
I have a matrix whose diagonal entries are positive whereas non-diagonal entries are negative.This matrix is also Strictly diagonally dominant.
Can we say that all elements of the inverse of this matrix is strictly positive i.e $a_{ij}$>0 .
linear-algebra matrices
linear-algebra matrices
asked Dec 30 '18 at 4:08
Ayush MishraAyush Mishra
485
asked Dec 30 '18 at 4:08
Ayush MishraAyush Mishra
485
edited Dec 30 '18 at 4:32
Ayush Mishra
asked Dec 30 '18 at 4:08
Ayush MishraAyush Mishra
485
asked Dec 30 '18 at 4:08
Ayush MishraAyush Mishra
485
asked Dec 30 '18 at 4:08
Ayush MishraAyush Mishra
485
485
$begingroup$
See en.wikipedia.org/wiki/M-matrix where it is said in the 3 first lines that it is a known result (but without proof)
$endgroup$
– Jean Marie
Dec 30 '18 at 18:41
add a comment |
$begingroup$
See en.wikipedia.org/wiki/M-matrix where it is said in the 3 first lines that it is a known result (but without proof)
$endgroup$
– Jean Marie
Dec 30 '18 at 18:41
$begingroup$
See en.wikipedia.org/wiki/M-matrix where it is said in the 3 first lines that it is a known result (but without proof)
$endgroup$
– Jean Marie
Dec 30 '18 at 18:41
$begingroup$
See en.wikipedia.org/wiki/M-matrix where it is said in the 3 first lines that it is a known result (but without proof)
$endgroup$
– Jean Marie
Dec 30 '18 at 18:41
add a comment |
1 Answer
1
active
oldest
votes
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Yes. Scale $A$ by a positive factor and we may assume that $max_ia_{ii}<1$. Then $B:=I-A$ is positive and
$$
sum_j|b_{ij}|
=|b_{ii}|+sum_{jne i}|b_{ij}|
=1-a_{ii}+sum_{jne i}|a_{ij}|
<1
$$
for each $i$. Hence $|B|_infty<1$ and we may expand $A^{-1}=(I-B)^{-1}$ as an infinite sum $I+B+B^2+ldots$. However, as $B$ is positive, the infinite sum is positive too.
answered Dec 30 '18 at 5:28
user1551user1551
72.1k566127
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Yes. Scale $A$ by a positive factor and we may assume that $max_ia_{ii}<1$. Then $B:=I-A$ is positive and
$$
sum_j|b_{ij}|
=|b_{ii}|+sum_{jne i}|b_{ij}|
=1-a_{ii}+sum_{jne i}|a_{ij}|
<1
$$
for each $i$. Hence $|B|_infty<1$ and we may expand $A^{-1}=(I-B)^{-1}$ as an infinite sum $I+B+B^2+ldots$. However, as $B$ is positive, the infinite sum is positive too.
answered Dec 30 '18 at 5:28
user1551user1551
72.1k566127
$endgroup$
add a comment |
$begingroup$
Yes. Scale $A$ by a positive factor and we may assume that $max_ia_{ii}<1$. Then $B:=I-A$ is positive and
$$
sum_j|b_{ij}|
=|b_{ii}|+sum_{jne i}|b_{ij}|
=1-a_{ii}+sum_{jne i}|a_{ij}|
<1
$$
for each $i$. Hence $|B|_infty<1$ and we may expand $A^{-1}=(I-B)^{-1}$ as an infinite sum $I+B+B^2+ldots$. However, as $B$ is positive, the infinite sum is positive too.
answered Dec 30 '18 at 5:28
user1551user1551
72.1k566127
$endgroup$
add a comment |
$begingroup$
Yes. Scale $A$ by a positive factor and we may assume that $max_ia_{ii}<1$. Then $B:=I-A$ is positive and
$$
sum_j|b_{ij}|
=|b_{ii}|+sum_{jne i}|b_{ij}|
=1-a_{ii}+sum_{jne i}|a_{ij}|
<1
$$
for each $i$. Hence $|B|_infty<1$ and we may expand $A^{-1}=(I-B)^{-1}$ as an infinite sum $I+B+B^2+ldots$. However, as $B$ is positive, the infinite sum is positive too.
answered Dec 30 '18 at 5:28
user1551user1551
72.1k566127
$endgroup$
Yes. Scale $A$ by a positive factor and we may assume that $max_ia_{ii}<1$. Then $B:=I-A$ is positive and
$$
sum_j|b_{ij}|
=|b_{ii}|+sum_{jne i}|b_{ij}|
=1-a_{ii}+sum_{jne i}|a_{ij}|
<1
$$
for each $i$. Hence $|B|_infty<1$ and we may expand $A^{-1}=(I-B)^{-1}$ as an infinite sum $I+B+B^2+ldots$. However, as $B$ is positive, the infinite sum is positive too.
answered Dec 30 '18 at 5:28
user1551user1551
72.1k566127
answered Dec 30 '18 at 5:28
user1551user1551
72.1k566127
answered Dec 30 '18 at 5:28
user1551user1551
72.1k566127
answered Dec 30 '18 at 5:28
user1551user1551
72.1k566127
72.1k566127
add a comment |
add a comment |
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$begingroup$
See en.wikipedia.org/wiki/M-matrix where it is said in the 3 first lines that it is a known result (but without proof)
$endgroup$
– Jean Marie
Dec 30 '18 at 18:41