Sparsity of the inverse of a non-negative matrix

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Let $A in mathbbR^b times b$ be a matrix with non-negative entries. Suppose that $A$ is invertible. Further suppose that all the diagonal entries of $A$ are non-null. My claim is the following:
$$textbin(A^-1)=textbin(A^b-1),,$$
where $textbin: mathbbR^b times brightarrow 0,1^b times b$ gives the binary sparsity pattern of $A$, in the sense that an entry of matrix $textbin(A)$ is $1$ if and only if the corresponding entry of $A$ is different from $0$. For instance
$$textbinleft(beginbmatrix1&2&0\0&2&1\1&0&2endbmatrix^-1right)=textbinleft(beginbmatrix1&2&0\0&2&1\1&0&2endbmatrix^2right)=beginbmatrix1&1&1\1&1&1\1&1&1endbmatrix,.$$



Notice that, by Cayley-Hamilton's theorem and the fact that the diagonal of $A$ is full, I expect any counter-example to my claim to be such that one entry of $textbin(A^-1)$ is $0$ while the same entry of $textbin(A^b-1)$ is $1$.




linear-algebra matrices inverse algebraic-graph-theory sparsity




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  • Sorry, I meant that there is an entry of $bin(A^-1)$ which is $0$ while the same entry of $bin(A^a-1)$ is $1$.
    – pulosky
    Jul 26 at 15:08










  • More concisely, you conjecture that if $A in mathbbR^a times a_geq 0$ is invertible, $A^-1$ has the same sparsity pattern as $A^a-1$. Moreover, you are asking if someone can find a counterexample in which $(A^-1)_ij = 0$ and $(A^a-1)_ij neq 0$ for some $(i, j)$.
    – parsiad
    Jul 26 at 15:13











  • Do you want $mboxbin(A^-1)$ to be equal to, less than or equal to, or greater than or equal to $mboxbin(A^b-1)$? You've described in this in different and inconsistent ways.
    – Brian Borchers
    Jul 26 at 15:18










  • @parsiad I additionally ask that $A$ has a full diagonal, so that the sparsity of $A$ "grows" with the powers of $A$. The counter-example I'm asking for is just a counter-example of the claim. at Rodrigo agreed. Changed it to $b times b$
    – pulosky
    Jul 26 at 15:19







  • 1




    That's helpful context- I'd suggest adding it to your question.
    – Brian Borchers
    Jul 26 at 15:26














up vote
2
down vote

favorite












Let $A in mathbbR^b times b$ be a matrix with non-negative entries. Suppose that $A$ is invertible. Further suppose that all the diagonal entries of $A$ are non-null. My claim is the following:
$$textbin(A^-1)=textbin(A^b-1),,$$
where $textbin: mathbbR^b times brightarrow 0,1^b times b$ gives the binary sparsity pattern of $A$, in the sense that an entry of matrix $textbin(A)$ is $1$ if and only if the corresponding entry of $A$ is different from $0$. For instance
$$textbinleft(beginbmatrix1&2&0\0&2&1\1&0&2endbmatrix^-1right)=textbinleft(beginbmatrix1&2&0\0&2&1\1&0&2endbmatrix^2right)=beginbmatrix1&1&1\1&1&1\1&1&1endbmatrix,.$$



Notice that, by Cayley-Hamilton's theorem and the fact that the diagonal of $A$ is full, I expect any counter-example to my claim to be such that one entry of $textbin(A^-1)$ is $0$ while the same entry of $textbin(A^b-1)$ is $1$.




linear-algebra matrices inverse algebraic-graph-theory sparsity




share|cite|improve this question





















  • Sorry, I meant that there is an entry of $bin(A^-1)$ which is $0$ while the same entry of $bin(A^a-1)$ is $1$.
    – pulosky
    Jul 26 at 15:08










  • More concisely, you conjecture that if $A in mathbbR^a times a_geq 0$ is invertible, $A^-1$ has the same sparsity pattern as $A^a-1$. Moreover, you are asking if someone can find a counterexample in which $(A^-1)_ij = 0$ and $(A^a-1)_ij neq 0$ for some $(i, j)$.
    – parsiad
    Jul 26 at 15:13











  • Do you want $mboxbin(A^-1)$ to be equal to, less than or equal to, or greater than or equal to $mboxbin(A^b-1)$? You've described in this in different and inconsistent ways.
    – Brian Borchers
    Jul 26 at 15:18










  • @parsiad I additionally ask that $A$ has a full diagonal, so that the sparsity of $A$ "grows" with the powers of $A$. The counter-example I'm asking for is just a counter-example of the claim. at Rodrigo agreed. Changed it to $b times b$
    – pulosky
    Jul 26 at 15:19







  • 1




    That's helpful context- I'd suggest adding it to your question.
    – Brian Borchers
    Jul 26 at 15:26












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $A in mathbbR^b times b$ be a matrix with non-negative entries. Suppose that $A$ is invertible. Further suppose that all the diagonal entries of $A$ are non-null. My claim is the following:
$$textbin(A^-1)=textbin(A^b-1),,$$
where $textbin: mathbbR^b times brightarrow 0,1^b times b$ gives the binary sparsity pattern of $A$, in the sense that an entry of matrix $textbin(A)$ is $1$ if and only if the corresponding entry of $A$ is different from $0$. For instance
$$textbinleft(beginbmatrix1&2&0\0&2&1\1&0&2endbmatrix^-1right)=textbinleft(beginbmatrix1&2&0\0&2&1\1&0&2endbmatrix^2right)=beginbmatrix1&1&1\1&1&1\1&1&1endbmatrix,.$$



Notice that, by Cayley-Hamilton's theorem and the fact that the diagonal of $A$ is full, I expect any counter-example to my claim to be such that one entry of $textbin(A^-1)$ is $0$ while the same entry of $textbin(A^b-1)$ is $1$.




linear-algebra matrices inverse algebraic-graph-theory sparsity




share|cite|improve this question













Let $A in mathbbR^b times b$ be a matrix with non-negative entries. Suppose that $A$ is invertible. Further suppose that all the diagonal entries of $A$ are non-null. My claim is the following:
$$textbin(A^-1)=textbin(A^b-1),,$$
where $textbin: mathbbR^b times brightarrow 0,1^b times b$ gives the binary sparsity pattern of $A$, in the sense that an entry of matrix $textbin(A)$ is $1$ if and only if the corresponding entry of $A$ is different from $0$. For instance
$$textbinleft(beginbmatrix1&2&0\0&2&1\1&0&2endbmatrix^-1right)=textbinleft(beginbmatrix1&2&0\0&2&1\1&0&2endbmatrix^2right)=beginbmatrix1&1&1\1&1&1\1&1&1endbmatrix,.$$



Notice that, by Cayley-Hamilton's theorem and the fact that the diagonal of $A$ is full, I expect any counter-example to my claim to be such that one entry of $textbin(A^-1)$ is $0$ while the same entry of $textbin(A^b-1)$ is $1$.





linear-algebra matrices inverse algebraic-graph-theory sparsity





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edited Jul 26 at 15:27
























asked Jul 26 at 14:52









pulosky

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  • Sorry, I meant that there is an entry of $bin(A^-1)$ which is $0$ while the same entry of $bin(A^a-1)$ is $1$.
    – pulosky
    Jul 26 at 15:08










  • More concisely, you conjecture that if $A in mathbbR^a times a_geq 0$ is invertible, $A^-1$ has the same sparsity pattern as $A^a-1$. Moreover, you are asking if someone can find a counterexample in which $(A^-1)_ij = 0$ and $(A^a-1)_ij neq 0$ for some $(i, j)$.
    – parsiad
    Jul 26 at 15:13











  • Do you want $mboxbin(A^-1)$ to be equal to, less than or equal to, or greater than or equal to $mboxbin(A^b-1)$? You've described in this in different and inconsistent ways.
    – Brian Borchers
    Jul 26 at 15:18










  • @parsiad I additionally ask that $A$ has a full diagonal, so that the sparsity of $A$ "grows" with the powers of $A$. The counter-example I'm asking for is just a counter-example of the claim. at Rodrigo agreed. Changed it to $b times b$
    – pulosky
    Jul 26 at 15:19







  • 1




    That's helpful context- I'd suggest adding it to your question.
    – Brian Borchers
    Jul 26 at 15:26
















  • Sorry, I meant that there is an entry of $bin(A^-1)$ which is $0$ while the same entry of $bin(A^a-1)$ is $1$.
    – pulosky
    Jul 26 at 15:08










  • More concisely, you conjecture that if $A in mathbbR^a times a_geq 0$ is invertible, $A^-1$ has the same sparsity pattern as $A^a-1$. Moreover, you are asking if someone can find a counterexample in which $(A^-1)_ij = 0$ and $(A^a-1)_ij neq 0$ for some $(i, j)$.
    – parsiad
    Jul 26 at 15:13











  • Do you want $mboxbin(A^-1)$ to be equal to, less than or equal to, or greater than or equal to $mboxbin(A^b-1)$? You've described in this in different and inconsistent ways.
    – Brian Borchers
    Jul 26 at 15:18










  • @parsiad I additionally ask that $A$ has a full diagonal, so that the sparsity of $A$ "grows" with the powers of $A$. The counter-example I'm asking for is just a counter-example of the claim. at Rodrigo agreed. Changed it to $b times b$
    – pulosky
    Jul 26 at 15:19







  • 1




    That's helpful context- I'd suggest adding it to your question.
    – Brian Borchers
    Jul 26 at 15:26















Sorry, I meant that there is an entry of $bin(A^-1)$ which is $0$ while the same entry of $bin(A^a-1)$ is $1$.
– pulosky
Jul 26 at 15:08




Sorry, I meant that there is an entry of $bin(A^-1)$ which is $0$ while the same entry of $bin(A^a-1)$ is $1$.
– pulosky
Jul 26 at 15:08












More concisely, you conjecture that if $A in mathbbR^a times a_geq 0$ is invertible, $A^-1$ has the same sparsity pattern as $A^a-1$. Moreover, you are asking if someone can find a counterexample in which $(A^-1)_ij = 0$ and $(A^a-1)_ij neq 0$ for some $(i, j)$.
– parsiad
Jul 26 at 15:13





More concisely, you conjecture that if $A in mathbbR^a times a_geq 0$ is invertible, $A^-1$ has the same sparsity pattern as $A^a-1$. Moreover, you are asking if someone can find a counterexample in which $(A^-1)_ij = 0$ and $(A^a-1)_ij neq 0$ for some $(i, j)$.
– parsiad
Jul 26 at 15:13













Do you want $mboxbin(A^-1)$ to be equal to, less than or equal to, or greater than or equal to $mboxbin(A^b-1)$? You've described in this in different and inconsistent ways.
– Brian Borchers
Jul 26 at 15:18




Do you want $mboxbin(A^-1)$ to be equal to, less than or equal to, or greater than or equal to $mboxbin(A^b-1)$? You've described in this in different and inconsistent ways.
– Brian Borchers
Jul 26 at 15:18












@parsiad I additionally ask that $A$ has a full diagonal, so that the sparsity of $A$ "grows" with the powers of $A$. The counter-example I'm asking for is just a counter-example of the claim. at Rodrigo agreed. Changed it to $b times b$
– pulosky
Jul 26 at 15:19





@parsiad I additionally ask that $A$ has a full diagonal, so that the sparsity of $A$ "grows" with the powers of $A$. The counter-example I'm asking for is just a counter-example of the claim. at Rodrigo agreed. Changed it to $b times b$
– pulosky
Jul 26 at 15:19





1




1




That's helpful context- I'd suggest adding it to your question.
– Brian Borchers
Jul 26 at 15:26




That's helpful context- I'd suggest adding it to your question.
– Brian Borchers
Jul 26 at 15:26










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accepted










$$ pmatrix1 & 1 & 1cr 1 & 2 & 2cr 1 & 2 & 3^-1 = pmatrix2 & -1 & 0cr
-1 & 2 & -1cr 0 & -1 & 1cr$$






share|cite|improve this answer





















  • great, thanks a lot!
    – pulosky
    Jul 26 at 15:29






  • 1




    @pulosky you can accept an answer to mark your question as answered
    – LinAlg
    Jul 26 at 15:53










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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










$$ pmatrix1 & 1 & 1cr 1 & 2 & 2cr 1 & 2 & 3^-1 = pmatrix2 & -1 & 0cr
-1 & 2 & -1cr 0 & -1 & 1cr$$






share|cite|improve this answer





















  • great, thanks a lot!
    – pulosky
    Jul 26 at 15:29






  • 1




    @pulosky you can accept an answer to mark your question as answered
    – LinAlg
    Jul 26 at 15:53














up vote
4
down vote



accepted










$$ pmatrix1 & 1 & 1cr 1 & 2 & 2cr 1 & 2 & 3^-1 = pmatrix2 & -1 & 0cr
-1 & 2 & -1cr 0 & -1 & 1cr$$






share|cite|improve this answer





















  • great, thanks a lot!
    – pulosky
    Jul 26 at 15:29






  • 1




    @pulosky you can accept an answer to mark your question as answered
    – LinAlg
    Jul 26 at 15:53












up vote
4
down vote



accepted







up vote
4
down vote



accepted






$$ pmatrix1 & 1 & 1cr 1 & 2 & 2cr 1 & 2 & 3^-1 = pmatrix2 & -1 & 0cr
-1 & 2 & -1cr 0 & -1 & 1cr$$






share|cite|improve this answer













$$ pmatrix1 & 1 & 1cr 1 & 2 & 2cr 1 & 2 & 3^-1 = pmatrix2 & -1 & 0cr
-1 & 2 & -1cr 0 & -1 & 1cr$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 26 at 15:27









Robert Israel

304k22201440




304k22201440











  • great, thanks a lot!
    – pulosky
    Jul 26 at 15:29






  • 1




    @pulosky you can accept an answer to mark your question as answered
    – LinAlg
    Jul 26 at 15:53
















  • great, thanks a lot!
    – pulosky
    Jul 26 at 15:29






  • 1




    @pulosky you can accept an answer to mark your question as answered
    – LinAlg
    Jul 26 at 15:53















great, thanks a lot!
– pulosky
Jul 26 at 15:29




great, thanks a lot!
– pulosky
Jul 26 at 15:29




1




1




@pulosky you can accept an answer to mark your question as answered
– LinAlg
Jul 26 at 15:53




@pulosky you can accept an answer to mark your question as answered
– LinAlg
Jul 26 at 15:53












 

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