Determinant of a specific symmetric matrix

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For matrix
$$
M_i,j = begincases
1-|a_i - a_j|,& textif a_ia_j>0 \
0, & textotherwise
endcases
$$
Can we prove that $det(M) ge 0$?
At least I could not find any counter-example numerically.
Any hint is also appreciated.



Thanks







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  • 1




    The diagonal entries are 1 if and only if $a_i neq 0$ for all i by your definition.
    – ertl
    Jul 28 at 15:40















up vote
3
down vote

favorite












For matrix
$$
M_i,j = begincases
1-|a_i - a_j|,& textif a_ia_j>0 \
0, & textotherwise
endcases
$$
Can we prove that $det(M) ge 0$?
At least I could not find any counter-example numerically.
Any hint is also appreciated.



Thanks







share|cite|improve this question

















  • 1




    The diagonal entries are 1 if and only if $a_i neq 0$ for all i by your definition.
    – ertl
    Jul 28 at 15:40













up vote
3
down vote

favorite









up vote
3
down vote

favorite











For matrix
$$
M_i,j = begincases
1-|a_i - a_j|,& textif a_ia_j>0 \
0, & textotherwise
endcases
$$
Can we prove that $det(M) ge 0$?
At least I could not find any counter-example numerically.
Any hint is also appreciated.



Thanks







share|cite|improve this question













For matrix
$$
M_i,j = begincases
1-|a_i - a_j|,& textif a_ia_j>0 \
0, & textotherwise
endcases
$$
Can we prove that $det(M) ge 0$?
At least I could not find any counter-example numerically.
Any hint is also appreciated.



Thanks









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday
























asked Jul 28 at 14:08









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1138







  • 1




    The diagonal entries are 1 if and only if $a_i neq 0$ for all i by your definition.
    – ertl
    Jul 28 at 15:40













  • 1




    The diagonal entries are 1 if and only if $a_i neq 0$ for all i by your definition.
    – ertl
    Jul 28 at 15:40








1




1




The diagonal entries are 1 if and only if $a_i neq 0$ for all i by your definition.
– ertl
Jul 28 at 15:40





The diagonal entries are 1 if and only if $a_i neq 0$ for all i by your definition.
– ertl
Jul 28 at 15:40











1 Answer
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It is easy to see that when the $a_i$s are arranged in descending order, $M$ becomes a direct sum of three diagonal sub-blocks, where the middle sub-block is zero and each of the other two sub-blocks is a smaller $M$ constructed with those nonzero $a_i$s of the same signs. Therefore it suffices to consider only the case where $a_i>0$ for each $i$.



Then $M$ is positive semidefinite because it is congruent to a nonnegative diagonal matrix. To illustrate, suppose $n=3$ and $a_1ge a_2ge a_3>0$. Rename $a_1,a_2,a_3$ as $a,b,c$. Then $M=E+A$, where $E$ denotes the all-one matrix and
$$
A=pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0.
$$
Therefore
beginalign
B&:=P^TAP\
&:=pmatrix1&-1\ &1&-1\ &&1
pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0
pmatrix1\ -1&1\ &-1&1\
&=pmatrixa-b&b-a&b-a\
b-c&b-c&c-b\
c-a&c-b&0
pmatrix1\ -1&1\ &-1&1\
&=pmatrix2(a-b)&0&b-a\
0&2(b-c)&c-b\
b-a&c-b&0,\
C&:=Q^TBQ\
&:=pmatrix1\ &1\ frac12&frac12&1
Bpmatrix1&&frac12\ &1&frac12\ &&1\
&=operatornamediagleft(2(a-b),2(b-c),0right).
endalign
Hence $Q^TP^TMPQ=operatornamediagleft(2(a-b),2(b-c),1right)$. Similarly, for a larger $n$, if we enlarge $P$ and $Q$ appropriately, we get
$$
Q^TP^TMPQ=operatornamediagleft(2(a_1-a_2),2(a_2-a_3),cdots,2(a_n-1-a_n),1right).
$$
Therefore $M$ is congruent to a nonnegative diagonal matrix and in turn it is positive semidefinite.






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    It is easy to see that when the $a_i$s are arranged in descending order, $M$ becomes a direct sum of three diagonal sub-blocks, where the middle sub-block is zero and each of the other two sub-blocks is a smaller $M$ constructed with those nonzero $a_i$s of the same signs. Therefore it suffices to consider only the case where $a_i>0$ for each $i$.



    Then $M$ is positive semidefinite because it is congruent to a nonnegative diagonal matrix. To illustrate, suppose $n=3$ and $a_1ge a_2ge a_3>0$. Rename $a_1,a_2,a_3$ as $a,b,c$. Then $M=E+A$, where $E$ denotes the all-one matrix and
    $$
    A=pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0.
    $$
    Therefore
    beginalign
    B&:=P^TAP\
    &:=pmatrix1&-1\ &1&-1\ &&1
    pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0
    pmatrix1\ -1&1\ &-1&1\
    &=pmatrixa-b&b-a&b-a\
    b-c&b-c&c-b\
    c-a&c-b&0
    pmatrix1\ -1&1\ &-1&1\
    &=pmatrix2(a-b)&0&b-a\
    0&2(b-c)&c-b\
    b-a&c-b&0,\
    C&:=Q^TBQ\
    &:=pmatrix1\ &1\ frac12&frac12&1
    Bpmatrix1&&frac12\ &1&frac12\ &&1\
    &=operatornamediagleft(2(a-b),2(b-c),0right).
    endalign
    Hence $Q^TP^TMPQ=operatornamediagleft(2(a-b),2(b-c),1right)$. Similarly, for a larger $n$, if we enlarge $P$ and $Q$ appropriately, we get
    $$
    Q^TP^TMPQ=operatornamediagleft(2(a_1-a_2),2(a_2-a_3),cdots,2(a_n-1-a_n),1right).
    $$
    Therefore $M$ is congruent to a nonnegative diagonal matrix and in turn it is positive semidefinite.






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      up vote
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      down vote













      It is easy to see that when the $a_i$s are arranged in descending order, $M$ becomes a direct sum of three diagonal sub-blocks, where the middle sub-block is zero and each of the other two sub-blocks is a smaller $M$ constructed with those nonzero $a_i$s of the same signs. Therefore it suffices to consider only the case where $a_i>0$ for each $i$.



      Then $M$ is positive semidefinite because it is congruent to a nonnegative diagonal matrix. To illustrate, suppose $n=3$ and $a_1ge a_2ge a_3>0$. Rename $a_1,a_2,a_3$ as $a,b,c$. Then $M=E+A$, where $E$ denotes the all-one matrix and
      $$
      A=pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0.
      $$
      Therefore
      beginalign
      B&:=P^TAP\
      &:=pmatrix1&-1\ &1&-1\ &&1
      pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0
      pmatrix1\ -1&1\ &-1&1\
      &=pmatrixa-b&b-a&b-a\
      b-c&b-c&c-b\
      c-a&c-b&0
      pmatrix1\ -1&1\ &-1&1\
      &=pmatrix2(a-b)&0&b-a\
      0&2(b-c)&c-b\
      b-a&c-b&0,\
      C&:=Q^TBQ\
      &:=pmatrix1\ &1\ frac12&frac12&1
      Bpmatrix1&&frac12\ &1&frac12\ &&1\
      &=operatornamediagleft(2(a-b),2(b-c),0right).
      endalign
      Hence $Q^TP^TMPQ=operatornamediagleft(2(a-b),2(b-c),1right)$. Similarly, for a larger $n$, if we enlarge $P$ and $Q$ appropriately, we get
      $$
      Q^TP^TMPQ=operatornamediagleft(2(a_1-a_2),2(a_2-a_3),cdots,2(a_n-1-a_n),1right).
      $$
      Therefore $M$ is congruent to a nonnegative diagonal matrix and in turn it is positive semidefinite.






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        It is easy to see that when the $a_i$s are arranged in descending order, $M$ becomes a direct sum of three diagonal sub-blocks, where the middle sub-block is zero and each of the other two sub-blocks is a smaller $M$ constructed with those nonzero $a_i$s of the same signs. Therefore it suffices to consider only the case where $a_i>0$ for each $i$.



        Then $M$ is positive semidefinite because it is congruent to a nonnegative diagonal matrix. To illustrate, suppose $n=3$ and $a_1ge a_2ge a_3>0$. Rename $a_1,a_2,a_3$ as $a,b,c$. Then $M=E+A$, where $E$ denotes the all-one matrix and
        $$
        A=pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0.
        $$
        Therefore
        beginalign
        B&:=P^TAP\
        &:=pmatrix1&-1\ &1&-1\ &&1
        pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0
        pmatrix1\ -1&1\ &-1&1\
        &=pmatrixa-b&b-a&b-a\
        b-c&b-c&c-b\
        c-a&c-b&0
        pmatrix1\ -1&1\ &-1&1\
        &=pmatrix2(a-b)&0&b-a\
        0&2(b-c)&c-b\
        b-a&c-b&0,\
        C&:=Q^TBQ\
        &:=pmatrix1\ &1\ frac12&frac12&1
        Bpmatrix1&&frac12\ &1&frac12\ &&1\
        &=operatornamediagleft(2(a-b),2(b-c),0right).
        endalign
        Hence $Q^TP^TMPQ=operatornamediagleft(2(a-b),2(b-c),1right)$. Similarly, for a larger $n$, if we enlarge $P$ and $Q$ appropriately, we get
        $$
        Q^TP^TMPQ=operatornamediagleft(2(a_1-a_2),2(a_2-a_3),cdots,2(a_n-1-a_n),1right).
        $$
        Therefore $M$ is congruent to a nonnegative diagonal matrix and in turn it is positive semidefinite.






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        It is easy to see that when the $a_i$s are arranged in descending order, $M$ becomes a direct sum of three diagonal sub-blocks, where the middle sub-block is zero and each of the other two sub-blocks is a smaller $M$ constructed with those nonzero $a_i$s of the same signs. Therefore it suffices to consider only the case where $a_i>0$ for each $i$.



        Then $M$ is positive semidefinite because it is congruent to a nonnegative diagonal matrix. To illustrate, suppose $n=3$ and $a_1ge a_2ge a_3>0$. Rename $a_1,a_2,a_3$ as $a,b,c$. Then $M=E+A$, where $E$ denotes the all-one matrix and
        $$
        A=pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0.
        $$
        Therefore
        beginalign
        B&:=P^TAP\
        &:=pmatrix1&-1\ &1&-1\ &&1
        pmatrix0&b-a&c-a\ b-a&0&c-b\ c-a&c-b&0
        pmatrix1\ -1&1\ &-1&1\
        &=pmatrixa-b&b-a&b-a\
        b-c&b-c&c-b\
        c-a&c-b&0
        pmatrix1\ -1&1\ &-1&1\
        &=pmatrix2(a-b)&0&b-a\
        0&2(b-c)&c-b\
        b-a&c-b&0,\
        C&:=Q^TBQ\
        &:=pmatrix1\ &1\ frac12&frac12&1
        Bpmatrix1&&frac12\ &1&frac12\ &&1\
        &=operatornamediagleft(2(a-b),2(b-c),0right).
        endalign
        Hence $Q^TP^TMPQ=operatornamediagleft(2(a-b),2(b-c),1right)$. Similarly, for a larger $n$, if we enlarge $P$ and $Q$ appropriately, we get
        $$
        Q^TP^TMPQ=operatornamediagleft(2(a_1-a_2),2(a_2-a_3),cdots,2(a_n-1-a_n),1right).
        $$
        Therefore $M$ is congruent to a nonnegative diagonal matrix and in turn it is positive semidefinite.







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        answered Jul 29 at 10:29









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