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Determinant of a specific symmetric matrix
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3
down vote
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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
matrices
asked Jul 28 at 14:08
989
1138
add a comment |Â
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
matrices
asked Jul 28 at 14:08
989
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
add a comment |Â
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
matrices
asked Jul 28 at 14:08
989
1138
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
matrices
asked Jul 28 at 14:08
989
1138
edited yesterday
asked Jul 28 at 14:08
989
1138
asked Jul 28 at 14:08
989
1138
asked Jul 28 at 14:08
989
1138
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
add a comment |Â
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
add a comment |Â
1 Answer
1
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up vote
1
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\ frac12½&1
Bpmatrix1&½\ &1½\ &&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.
answered Jul 29 at 10:29
user1551
66.8k564122
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
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\ frac12½&1
Bpmatrix1&½\ &1½\ &&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.
answered Jul 29 at 10:29
user1551
66.8k564122
add a comment |Â
up vote
1
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\ frac12½&1
Bpmatrix1&½\ &1½\ &&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.
answered Jul 29 at 10:29
user1551
66.8k564122
add a comment |Â
up vote
1
down vote
up vote
1
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\ frac12½&1
Bpmatrix1&½\ &1½\ &&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.
answered Jul 29 at 10:29
user1551
66.8k564122
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\ frac12½&1
Bpmatrix1&½\ &1½\ &&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.
answered Jul 29 at 10:29
user1551
66.8k564122
answered Jul 29 at 10:29
user1551
66.8k564122
answered Jul 29 at 10:29
user1551
66.8k564122
answered Jul 29 at 10:29
user1551
66.8k564122
66.8k564122
add a comment |Â
add a comment |Â
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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