Mixing
Element-wise ordering of the inverse of two M-matrices
$begingroup$
Assume $A$ and $B$ are two M-mtrices and $A geq B$, where "$geq$" is element-wise ordering. Can I show that $A^{-1} leq B^{-1}$?
I think it holds if $D = A - B$ does not have all-zero rows/columns.
linear-algebra matrices inverse
asked Jan 26 at 2:16
AmirAmir
858
$endgroup$
add a comment |
$begingroup$
Assume $A$ and $B$ are two M-mtrices and $A geq B$, where "$geq$" is element-wise ordering. Can I show that $A^{-1} leq B^{-1}$?
I think it holds if $D = A - B$ does not have all-zero rows/columns.
linear-algebra matrices inverse
asked Jan 26 at 2:16
AmirAmir
858
$endgroup$
1
$begingroup$
What are "M-matrices"?
$endgroup$
– coffeemath
Jan 26 at 2:20
$begingroup$
en.wikipedia.org/wiki/M-matrix
$endgroup$
– Amir
Jan 26 at 2:22
$begingroup$
The order implied by the post and the implied proposition of the title don't agree. Should the off-diagonal elements of $A-B$ be positive or negative?
$endgroup$
– jmerry
Jan 26 at 5:01
$begingroup$
I updated the title, @jmerry.
$endgroup$
– Amir
Jan 27 at 21:04
add a comment |
$begingroup$
Assume $A$ and $B$ are two M-mtrices and $A geq B$, where "$geq$" is element-wise ordering. Can I show that $A^{-1} leq B^{-1}$?
I think it holds if $D = A - B$ does not have all-zero rows/columns.
linear-algebra matrices inverse
asked Jan 26 at 2:16
AmirAmir
858
$endgroup$
Assume $A$ and $B$ are two M-mtrices and $A geq B$, where "$geq$" is element-wise ordering. Can I show that $A^{-1} leq B^{-1}$?
I think it holds if $D = A - B$ does not have all-zero rows/columns.
linear-algebra matrices inverse
linear-algebra matrices inverse
asked Jan 26 at 2:16
AmirAmir
858
asked Jan 26 at 2:16
AmirAmir
858
edited Jan 28 at 13:42
Amir
asked Jan 26 at 2:16
AmirAmir
858
asked Jan 26 at 2:16
AmirAmir
858
asked Jan 26 at 2:16
AmirAmir
858
858
1
$begingroup$
What are "M-matrices"?
$endgroup$
– coffeemath
Jan 26 at 2:20
$begingroup$
en.wikipedia.org/wiki/M-matrix
$endgroup$
– Amir
Jan 26 at 2:22
$begingroup$
The order implied by the post and the implied proposition of the title don't agree. Should the off-diagonal elements of $A-B$ be positive or negative?
$endgroup$
– jmerry
Jan 26 at 5:01
$begingroup$
I updated the title, @jmerry.
$endgroup$
– Amir
Jan 27 at 21:04
add a comment |
1
$begingroup$
What are "M-matrices"?
$endgroup$
– coffeemath
Jan 26 at 2:20
$begingroup$
en.wikipedia.org/wiki/M-matrix
$endgroup$
– Amir
Jan 26 at 2:22
$begingroup$
The order implied by the post and the implied proposition of the title don't agree. Should the off-diagonal elements of $A-B$ be positive or negative?
$endgroup$
– jmerry
Jan 26 at 5:01
$begingroup$
I updated the title, @jmerry.
$endgroup$
– Amir
Jan 27 at 21:04
1
1
$begingroup$
What are "M-matrices"?
$endgroup$
– coffeemath
Jan 26 at 2:20
$begingroup$
What are "M-matrices"?
$endgroup$
– coffeemath
Jan 26 at 2:20
$begingroup$
en.wikipedia.org/wiki/M-matrix
$endgroup$
– Amir
Jan 26 at 2:22
$begingroup$
en.wikipedia.org/wiki/M-matrix
$endgroup$
– Amir
Jan 26 at 2:22
$begingroup$
The order implied by the post and the implied proposition of the title don't agree. Should the off-diagonal elements of $A-B$ be positive or negative?
$endgroup$
– jmerry
Jan 26 at 5:01
$begingroup$
The order implied by the post and the implied proposition of the title don't agree. Should the off-diagonal elements of $A-B$ be positive or negative?
$endgroup$
– jmerry
Jan 26 at 5:01
$begingroup$
I updated the title, @jmerry.
$endgroup$
– Amir
Jan 27 at 21:04
$begingroup$
I updated the title, @jmerry.
$endgroup$
– Amir
Jan 27 at 21:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$A$ has a positive diagonal because it is a nonsingular M-matrix. As invertibility, entrywise order and inverse-positivity of matrices are all preserved under congruence by positive diagonal matrix, we may assume that all diagonal entries of $A$ are equal to $1$. With this additional assumption, we may write $A=I-S$, where $S$ is a nonnegative matrix with a zero diagonal and $rho(S)<1$.
Now, as $B$ is a nonsingular M-matrix that is $le A$, it can be written as $bI-T$ where $rho(T)<b:=max_ib_{ii}le1$ and $Tge S$ (this is where we need $A$ to have a constant diagonal: without this assumption, we cannot force the diagonal of $S$ to zero and in turn we cannot guarantee that the diagonal part of $S$ is dominated by the diagonal of $T$, even though it is assumed that $Age B$). But then $frac1{b^{k+1}}T^k ge T^kge S^k$ for every nonnegative integer $k$. Therefore
begin{aligned}
A^{-1}=(I-S)^{-1}
&=I+S+S^2+ldots\
&lefrac1bleft(I+frac1bT+frac1{b^2}T^2+ldotsright)\
&=frac1bleft(I-frac1bTright)^{-1}=B^{-1}.
end{aligned}
answered Jan 28 at 15:40
user1551user1551
73.7k566129
$endgroup$
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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oldest
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active
oldest
votes
$begingroup$
$A$ has a positive diagonal because it is a nonsingular M-matrix. As invertibility, entrywise order and inverse-positivity of matrices are all preserved under congruence by positive diagonal matrix, we may assume that all diagonal entries of $A$ are equal to $1$. With this additional assumption, we may write $A=I-S$, where $S$ is a nonnegative matrix with a zero diagonal and $rho(S)<1$.
Now, as $B$ is a nonsingular M-matrix that is $le A$, it can be written as $bI-T$ where $rho(T)<b:=max_ib_{ii}le1$ and $Tge S$ (this is where we need $A$ to have a constant diagonal: without this assumption, we cannot force the diagonal of $S$ to zero and in turn we cannot guarantee that the diagonal part of $S$ is dominated by the diagonal of $T$, even though it is assumed that $Age B$). But then $frac1{b^{k+1}}T^k ge T^kge S^k$ for every nonnegative integer $k$. Therefore
begin{aligned}
A^{-1}=(I-S)^{-1}
&=I+S+S^2+ldots\
&lefrac1bleft(I+frac1bT+frac1{b^2}T^2+ldotsright)\
&=frac1bleft(I-frac1bTright)^{-1}=B^{-1}.
end{aligned}
answered Jan 28 at 15:40
user1551user1551
73.7k566129
$endgroup$
add a comment |
$begingroup$
$A$ has a positive diagonal because it is a nonsingular M-matrix. As invertibility, entrywise order and inverse-positivity of matrices are all preserved under congruence by positive diagonal matrix, we may assume that all diagonal entries of $A$ are equal to $1$. With this additional assumption, we may write $A=I-S$, where $S$ is a nonnegative matrix with a zero diagonal and $rho(S)<1$.
Now, as $B$ is a nonsingular M-matrix that is $le A$, it can be written as $bI-T$ where $rho(T)<b:=max_ib_{ii}le1$ and $Tge S$ (this is where we need $A$ to have a constant diagonal: without this assumption, we cannot force the diagonal of $S$ to zero and in turn we cannot guarantee that the diagonal part of $S$ is dominated by the diagonal of $T$, even though it is assumed that $Age B$). But then $frac1{b^{k+1}}T^k ge T^kge S^k$ for every nonnegative integer $k$. Therefore
begin{aligned}
A^{-1}=(I-S)^{-1}
&=I+S+S^2+ldots\
&lefrac1bleft(I+frac1bT+frac1{b^2}T^2+ldotsright)\
&=frac1bleft(I-frac1bTright)^{-1}=B^{-1}.
end{aligned}
answered Jan 28 at 15:40
user1551user1551
73.7k566129
$endgroup$
add a comment |
$begingroup$
$A$ has a positive diagonal because it is a nonsingular M-matrix. As invertibility, entrywise order and inverse-positivity of matrices are all preserved under congruence by positive diagonal matrix, we may assume that all diagonal entries of $A$ are equal to $1$. With this additional assumption, we may write $A=I-S$, where $S$ is a nonnegative matrix with a zero diagonal and $rho(S)<1$.
Now, as $B$ is a nonsingular M-matrix that is $le A$, it can be written as $bI-T$ where $rho(T)<b:=max_ib_{ii}le1$ and $Tge S$ (this is where we need $A$ to have a constant diagonal: without this assumption, we cannot force the diagonal of $S$ to zero and in turn we cannot guarantee that the diagonal part of $S$ is dominated by the diagonal of $T$, even though it is assumed that $Age B$). But then $frac1{b^{k+1}}T^k ge T^kge S^k$ for every nonnegative integer $k$. Therefore
begin{aligned}
A^{-1}=(I-S)^{-1}
&=I+S+S^2+ldots\
&lefrac1bleft(I+frac1bT+frac1{b^2}T^2+ldotsright)\
&=frac1bleft(I-frac1bTright)^{-1}=B^{-1}.
end{aligned}
answered Jan 28 at 15:40
user1551user1551
73.7k566129
$endgroup$
$A$ has a positive diagonal because it is a nonsingular M-matrix. As invertibility, entrywise order and inverse-positivity of matrices are all preserved under congruence by positive diagonal matrix, we may assume that all diagonal entries of $A$ are equal to $1$. With this additional assumption, we may write $A=I-S$, where $S$ is a nonnegative matrix with a zero diagonal and $rho(S)<1$.
Now, as $B$ is a nonsingular M-matrix that is $le A$, it can be written as $bI-T$ where $rho(T)<b:=max_ib_{ii}le1$ and $Tge S$ (this is where we need $A$ to have a constant diagonal: without this assumption, we cannot force the diagonal of $S$ to zero and in turn we cannot guarantee that the diagonal part of $S$ is dominated by the diagonal of $T$, even though it is assumed that $Age B$). But then $frac1{b^{k+1}}T^k ge T^kge S^k$ for every nonnegative integer $k$. Therefore
begin{aligned}
A^{-1}=(I-S)^{-1}
&=I+S+S^2+ldots\
&lefrac1bleft(I+frac1bT+frac1{b^2}T^2+ldotsright)\
&=frac1bleft(I-frac1bTright)^{-1}=B^{-1}.
end{aligned}
answered Jan 28 at 15:40
user1551user1551
73.7k566129
answered Jan 28 at 15:40
user1551user1551
73.7k566129
answered Jan 28 at 15:40
user1551user1551
73.7k566129
answered Jan 28 at 15:40
user1551user1551
73.7k566129
73.7k566129
add a comment |
add a comment |
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1
$begingroup$
What are "M-matrices"?
$endgroup$
– coffeemath
Jan 26 at 2:20
$begingroup$
en.wikipedia.org/wiki/M-matrix
$endgroup$
– Amir
Jan 26 at 2:22
$begingroup$
The order implied by the post and the implied proposition of the title don't agree. Should the off-diagonal elements of $A-B$ be positive or negative?
$endgroup$
– jmerry
Jan 26 at 5:01
$begingroup$
I updated the title, @jmerry.
$endgroup$
– Amir
Jan 27 at 21:04