Mixing
equivalence of two conditions on non-negative square matrix
$begingroup$
Let $A$ be a nonnegative square matrix (every entry is non-negative, $Age 0$). Look at these two conditions:
(1) $A^nto0$ as $ntoinfty$
(2) $(I-A)^{-1}ge 0$
I can show that (1) implies (2). In this case, the inverse is $I+A+A^2+dots$.
My question is
"Is the converse true? i.e. does (2) imply (1) as well?"
linear-algebra inverse economics
asked Jan 29 at 3:05
Tony BTony B
822518
$endgroup$
add a comment |
$begingroup$
Let $A$ be a nonnegative square matrix (every entry is non-negative, $Age 0$). Look at these two conditions:
(1) $A^nto0$ as $ntoinfty$
(2) $(I-A)^{-1}ge 0$
I can show that (1) implies (2). In this case, the inverse is $I+A+A^2+dots$.
My question is
"Is the converse true? i.e. does (2) imply (1) as well?"
linear-algebra inverse economics
asked Jan 29 at 3:05
Tony BTony B
822518
$endgroup$
1
$begingroup$
When you say "non-negative" square matrix, do you mean that all entries are non-negative, that the matrix is positive semi-definite, or something else?
$endgroup$
– Theo Bendit
Jan 29 at 3:09
$begingroup$
I think he meant the determinant is not negative. Not sure.
$endgroup$
– Neo Darwin
Jan 29 at 3:11
$begingroup$
@TheoBendit All entries are non-negative.
$endgroup$
– Tony B
Jan 29 at 3:59
add a comment |
$begingroup$
Let $A$ be a nonnegative square matrix (every entry is non-negative, $Age 0$). Look at these two conditions:
(1) $A^nto0$ as $ntoinfty$
(2) $(I-A)^{-1}ge 0$
I can show that (1) implies (2). In this case, the inverse is $I+A+A^2+dots$.
My question is
"Is the converse true? i.e. does (2) imply (1) as well?"
linear-algebra inverse economics
asked Jan 29 at 3:05
Tony BTony B
822518
$endgroup$
Let $A$ be a nonnegative square matrix (every entry is non-negative, $Age 0$). Look at these two conditions:
(1) $A^nto0$ as $ntoinfty$
(2) $(I-A)^{-1}ge 0$
I can show that (1) implies (2). In this case, the inverse is $I+A+A^2+dots$.
My question is
"Is the converse true? i.e. does (2) imply (1) as well?"
linear-algebra inverse economics
linear-algebra inverse economics
asked Jan 29 at 3:05
Tony BTony B
822518
asked Jan 29 at 3:05
Tony BTony B
822518
edited Jan 29 at 3:59
Tony B
asked Jan 29 at 3:05
Tony BTony B
822518
asked Jan 29 at 3:05
Tony BTony B
822518
asked Jan 29 at 3:05
Tony BTony B
822518
822518
1
$begingroup$
When you say "non-negative" square matrix, do you mean that all entries are non-negative, that the matrix is positive semi-definite, or something else?
$endgroup$
– Theo Bendit
Jan 29 at 3:09
$begingroup$
I think he meant the determinant is not negative. Not sure.
$endgroup$
– Neo Darwin
Jan 29 at 3:11
$begingroup$
@TheoBendit All entries are non-negative.
$endgroup$
– Tony B
Jan 29 at 3:59
add a comment |
1
$begingroup$
When you say "non-negative" square matrix, do you mean that all entries are non-negative, that the matrix is positive semi-definite, or something else?
$endgroup$
– Theo Bendit
Jan 29 at 3:09
$begingroup$
I think he meant the determinant is not negative. Not sure.
$endgroup$
– Neo Darwin
Jan 29 at 3:11
$begingroup$
@TheoBendit All entries are non-negative.
$endgroup$
– Tony B
Jan 29 at 3:59
1
1
$begingroup$
When you say "non-negative" square matrix, do you mean that all entries are non-negative, that the matrix is positive semi-definite, or something else?
$endgroup$
– Theo Bendit
Jan 29 at 3:09
$begingroup$
When you say "non-negative" square matrix, do you mean that all entries are non-negative, that the matrix is positive semi-definite, or something else?
$endgroup$
– Theo Bendit
Jan 29 at 3:09
$begingroup$
I think he meant the determinant is not negative. Not sure.
$endgroup$
– Neo Darwin
Jan 29 at 3:11
$begingroup$
I think he meant the determinant is not negative. Not sure.
$endgroup$
– Neo Darwin
Jan 29 at 3:11
$begingroup$
@TheoBendit All entries are non-negative.
$endgroup$
– Tony B
Jan 29 at 3:59
$begingroup$
@TheoBendit All entries are non-negative.
$endgroup$
– Tony B
Jan 29 at 3:59
add a comment |
1 Answer
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Yes. Let $x$ be the Perron vector of $A$. Then $0le(I-A)^{-1}x=(1-rho(A))^{-1}x$. Hence $(1-rho(A))^{-1}>0$ and $rho(A)<1$. In turn, $A^nto0$ as $ntoinfty$.
answered Jan 29 at 4:39
user1551user1551
73.9k566129
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$begingroup$
Yes. Let $x$ be the Perron vector of $A$. Then $0le(I-A)^{-1}x=(1-rho(A))^{-1}x$. Hence $(1-rho(A))^{-1}>0$ and $rho(A)<1$. In turn, $A^nto0$ as $ntoinfty$.
answered Jan 29 at 4:39
user1551user1551
73.9k566129
$endgroup$
add a comment |
$begingroup$
Yes. Let $x$ be the Perron vector of $A$. Then $0le(I-A)^{-1}x=(1-rho(A))^{-1}x$. Hence $(1-rho(A))^{-1}>0$ and $rho(A)<1$. In turn, $A^nto0$ as $ntoinfty$.
answered Jan 29 at 4:39
user1551user1551
73.9k566129
$endgroup$
add a comment |
$begingroup$
Yes. Let $x$ be the Perron vector of $A$. Then $0le(I-A)^{-1}x=(1-rho(A))^{-1}x$. Hence $(1-rho(A))^{-1}>0$ and $rho(A)<1$. In turn, $A^nto0$ as $ntoinfty$.
answered Jan 29 at 4:39
user1551user1551
73.9k566129
$endgroup$
Yes. Let $x$ be the Perron vector of $A$. Then $0le(I-A)^{-1}x=(1-rho(A))^{-1}x$. Hence $(1-rho(A))^{-1}>0$ and $rho(A)<1$. In turn, $A^nto0$ as $ntoinfty$.
answered Jan 29 at 4:39
user1551user1551
73.9k566129
answered Jan 29 at 4:39
user1551user1551
73.9k566129
answered Jan 29 at 4:39
user1551user1551
73.9k566129
answered Jan 29 at 4:39
user1551user1551
73.9k566129
73.9k566129
add a comment |
add a comment |
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1
$begingroup$
When you say "non-negative" square matrix, do you mean that all entries are non-negative, that the matrix is positive semi-definite, or something else?
$endgroup$
– Theo Bendit
Jan 29 at 3:09
$begingroup$
I think he meant the determinant is not negative. Not sure.
$endgroup$
– Neo Darwin
Jan 29 at 3:11
$begingroup$
@TheoBendit All entries are non-negative.
$endgroup$
– Tony B
Jan 29 at 3:59