Mixing
non-negative matrix satisfying two conditions
$begingroup$
A real matrix $B$ is called non-negative if every entry is non-negative. We will denote this by $Bge 0$.
I want to find a non-negative matrix $B$ satisfying the following two conditions:
(1) $(I-B)^{-1}$ exists but not non-negative. Here $I$ is the identity matrix.
(2) There is a non-zero and non-negative vector $vec{d}$ such that $(I-B)^{-1}vec{d}ge 0$.
I tried all the $2times 2$ matrices, but it did not work. I conjecture that such a $B$ does not exist, but don't know how to prove it.
linear-algebra eigenvalues-eigenvectors inverse economics
asked Jan 29 at 16:23
Tony BTony B
822518
$endgroup$
add a comment |
$begingroup$
A real matrix $B$ is called non-negative if every entry is non-negative. We will denote this by $Bge 0$.
I want to find a non-negative matrix $B$ satisfying the following two conditions:
(1) $(I-B)^{-1}$ exists but not non-negative. Here $I$ is the identity matrix.
(2) There is a non-zero and non-negative vector $vec{d}$ such that $(I-B)^{-1}vec{d}ge 0$.
I tried all the $2times 2$ matrices, but it did not work. I conjecture that such a $B$ does not exist, but don't know how to prove it.
linear-algebra eigenvalues-eigenvectors inverse economics
asked Jan 29 at 16:23
Tony BTony B
822518
$endgroup$
$begingroup$
There are conditions under which $(I-B)^{-1}=I+B+B^2+ldots$. In that case it's clear that $(I-B)^{-1}$ would be non-negative. Any counterexample would have be such that $(I-B)^{-1} neq I+B+B^{2}+ldots$.
$endgroup$
– Brian Borchers
Jan 29 at 16:29
$begingroup$
@BrianBorchers I agree. $B^n$ can not converge to $0$
$endgroup$
– Tony B
Jan 29 at 18:32
$begingroup$
So what happens if $B$ is a really big nonnegative matrix?
$endgroup$
– Brian Borchers
Jan 29 at 19:44
add a comment |
$begingroup$
A real matrix $B$ is called non-negative if every entry is non-negative. We will denote this by $Bge 0$.
I want to find a non-negative matrix $B$ satisfying the following two conditions:
(1) $(I-B)^{-1}$ exists but not non-negative. Here $I$ is the identity matrix.
(2) There is a non-zero and non-negative vector $vec{d}$ such that $(I-B)^{-1}vec{d}ge 0$.
I tried all the $2times 2$ matrices, but it did not work. I conjecture that such a $B$ does not exist, but don't know how to prove it.
linear-algebra eigenvalues-eigenvectors inverse economics
asked Jan 29 at 16:23
Tony BTony B
822518
$endgroup$
A real matrix $B$ is called non-negative if every entry is non-negative. We will denote this by $Bge 0$.
I want to find a non-negative matrix $B$ satisfying the following two conditions:
(1) $(I-B)^{-1}$ exists but not non-negative. Here $I$ is the identity matrix.
(2) There is a non-zero and non-negative vector $vec{d}$ such that $(I-B)^{-1}vec{d}ge 0$.
I tried all the $2times 2$ matrices, but it did not work. I conjecture that such a $B$ does not exist, but don't know how to prove it.
linear-algebra eigenvalues-eigenvectors inverse economics
linear-algebra eigenvalues-eigenvectors inverse economics
asked Jan 29 at 16:23
Tony BTony B
822518
asked Jan 29 at 16:23
Tony BTony B
822518
asked Jan 29 at 16:23
Tony BTony B
822518
asked Jan 29 at 16:23
Tony BTony B
822518
asked Jan 29 at 16:23
Tony BTony B
822518
822518
$begingroup$
There are conditions under which $(I-B)^{-1}=I+B+B^2+ldots$. In that case it's clear that $(I-B)^{-1}$ would be non-negative. Any counterexample would have be such that $(I-B)^{-1} neq I+B+B^{2}+ldots$.
$endgroup$
– Brian Borchers
Jan 29 at 16:29
$begingroup$
@BrianBorchers I agree. $B^n$ can not converge to $0$
$endgroup$
– Tony B
Jan 29 at 18:32
$begingroup$
So what happens if $B$ is a really big nonnegative matrix?
$endgroup$
– Brian Borchers
Jan 29 at 19:44
add a comment |
$begingroup$
There are conditions under which $(I-B)^{-1}=I+B+B^2+ldots$. In that case it's clear that $(I-B)^{-1}$ would be non-negative. Any counterexample would have be such that $(I-B)^{-1} neq I+B+B^{2}+ldots$.
$endgroup$
– Brian Borchers
Jan 29 at 16:29
$begingroup$
@BrianBorchers I agree. $B^n$ can not converge to $0$
$endgroup$
– Tony B
Jan 29 at 18:32
$begingroup$
So what happens if $B$ is a really big nonnegative matrix?
$endgroup$
– Brian Borchers
Jan 29 at 19:44
$begingroup$
There are conditions under which $(I-B)^{-1}=I+B+B^2+ldots$. In that case it's clear that $(I-B)^{-1}$ would be non-negative. Any counterexample would have be such that $(I-B)^{-1} neq I+B+B^{2}+ldots$.
$endgroup$
– Brian Borchers
Jan 29 at 16:29
$begingroup$
There are conditions under which $(I-B)^{-1}=I+B+B^2+ldots$. In that case it's clear that $(I-B)^{-1}$ would be non-negative. Any counterexample would have be such that $(I-B)^{-1} neq I+B+B^{2}+ldots$.
$endgroup$
– Brian Borchers
Jan 29 at 16:29
$begingroup$
@BrianBorchers I agree. $B^n$ can not converge to $0$
$endgroup$
– Tony B
Jan 29 at 18:32
$begingroup$
@BrianBorchers I agree. $B^n$ can not converge to $0$
$endgroup$
– Tony B
Jan 29 at 18:32
$begingroup$
So what happens if $B$ is a really big nonnegative matrix?
$endgroup$
– Brian Borchers
Jan 29 at 19:44
$begingroup$
So what happens if $B$ is a really big nonnegative matrix?
$endgroup$
– Brian Borchers
Jan 29 at 19:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For example $B=diag(1/2,2)$; then $(I-B)^{-1}=diag(2,-1)$ and we can choose $d=[1,0]^T$.
Do you understand why such a $B$ works ?
answered Jan 30 at 21:09
loup blancloup blanc
24.1k21851
$endgroup$
$begingroup$
The inverse should be diag(2,-1). But I got your point. Your example works! From the economics point of view (input-output model), one industry is profitable, and the other is not. They don't interact with each other.
$endgroup$
– Tony B
Jan 31 at 22:20
$begingroup$
Yes, exactly. On the other hand, I go to correct the miscalculation.
$endgroup$
– loup blanc
Jan 31 at 22:23
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
For example $B=diag(1/2,2)$; then $(I-B)^{-1}=diag(2,-1)$ and we can choose $d=[1,0]^T$.
Do you understand why such a $B$ works ?
answered Jan 30 at 21:09
loup blancloup blanc
24.1k21851
$endgroup$
$begingroup$
The inverse should be diag(2,-1). But I got your point. Your example works! From the economics point of view (input-output model), one industry is profitable, and the other is not. They don't interact with each other.
$endgroup$
– Tony B
Jan 31 at 22:20
$begingroup$
Yes, exactly. On the other hand, I go to correct the miscalculation.
$endgroup$
– loup blanc
Jan 31 at 22:23
add a comment |
$begingroup$
For example $B=diag(1/2,2)$; then $(I-B)^{-1}=diag(2,-1)$ and we can choose $d=[1,0]^T$.
Do you understand why such a $B$ works ?
answered Jan 30 at 21:09
loup blancloup blanc
24.1k21851
$endgroup$
$begingroup$
The inverse should be diag(2,-1). But I got your point. Your example works! From the economics point of view (input-output model), one industry is profitable, and the other is not. They don't interact with each other.
$endgroup$
– Tony B
Jan 31 at 22:20
$begingroup$
Yes, exactly. On the other hand, I go to correct the miscalculation.
$endgroup$
– loup blanc
Jan 31 at 22:23
add a comment |
$begingroup$
For example $B=diag(1/2,2)$; then $(I-B)^{-1}=diag(2,-1)$ and we can choose $d=[1,0]^T$.
Do you understand why such a $B$ works ?
answered Jan 30 at 21:09
loup blancloup blanc
24.1k21851
$endgroup$
For example $B=diag(1/2,2)$; then $(I-B)^{-1}=diag(2,-1)$ and we can choose $d=[1,0]^T$.
Do you understand why such a $B$ works ?
answered Jan 30 at 21:09
loup blancloup blanc
24.1k21851
edited Jan 31 at 22:23
answered Jan 30 at 21:09
loup blancloup blanc
24.1k21851
answered Jan 30 at 21:09
loup blancloup blanc
24.1k21851
answered Jan 30 at 21:09
loup blancloup blanc
24.1k21851
24.1k21851
$begingroup$
The inverse should be diag(2,-1). But I got your point. Your example works! From the economics point of view (input-output model), one industry is profitable, and the other is not. They don't interact with each other.
$endgroup$
– Tony B
Jan 31 at 22:20
$begingroup$
Yes, exactly. On the other hand, I go to correct the miscalculation.
$endgroup$
– loup blanc
Jan 31 at 22:23
add a comment |
$begingroup$
The inverse should be diag(2,-1). But I got your point. Your example works! From the economics point of view (input-output model), one industry is profitable, and the other is not. They don't interact with each other.
$endgroup$
– Tony B
Jan 31 at 22:20
$begingroup$
Yes, exactly. On the other hand, I go to correct the miscalculation.
$endgroup$
– loup blanc
Jan 31 at 22:23
$begingroup$
The inverse should be diag(2,-1). But I got your point. Your example works! From the economics point of view (input-output model), one industry is profitable, and the other is not. They don't interact with each other.
$endgroup$
– Tony B
Jan 31 at 22:20
$begingroup$
The inverse should be diag(2,-1). But I got your point. Your example works! From the economics point of view (input-output model), one industry is profitable, and the other is not. They don't interact with each other.
$endgroup$
– Tony B
Jan 31 at 22:20
$begingroup$
Yes, exactly. On the other hand, I go to correct the miscalculation.
$endgroup$
– loup blanc
Jan 31 at 22:23
$begingroup$
Yes, exactly. On the other hand, I go to correct the miscalculation.
$endgroup$
– loup blanc
Jan 31 at 22:23
add a comment |
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$begingroup$
There are conditions under which $(I-B)^{-1}=I+B+B^2+ldots$. In that case it's clear that $(I-B)^{-1}$ would be non-negative. Any counterexample would have be such that $(I-B)^{-1} neq I+B+B^{2}+ldots$.
$endgroup$
– Brian Borchers
Jan 29 at 16:29
$begingroup$
@BrianBorchers I agree. $B^n$ can not converge to $0$
$endgroup$
– Tony B
Jan 29 at 18:32
$begingroup$
So what happens if $B$ is a really big nonnegative matrix?
$endgroup$
– Brian Borchers
Jan 29 at 19:44