Rewrite a condition on a $3times3$ matrixWhich of the following statements are falseColumn space of complex matrixFinding determinant of $n times n$ matrixGuass Jordan Elimination Matrix ProblemEvaluate the following determinant:To show: $beginvmatrix -bc & b^2+bc & c^2+bc\ a^2+ac & -ac & c^2+ac \ a^2+ab & b^2+ab & -ab endvmatrix= (ab+bc+ca)^3$if $p+q+r=0$ Find the value of the DeterminantWhat is the rank of a matrix when the difference between its rows is same?Determinant of beginvmatrix -2a &a+b &a+c \ b+a& -2b &b+c \ c+a&c+b & -2c endvmatrixDefined row operations for making a matrix ,containing varibel coefficients, into RREF
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Rewrite a condition on a $3times3$ matrix
Which of the following statements are falseColumn space of complex matrixFinding determinant of $n times n$ matrixGuass Jordan Elimination Matrix ProblemEvaluate the following determinant:To show: $beginvmatrix -bc & b^2+bc & c^2+bc\ a^2+ac & -ac & c^2+ac \ a^2+ab & b^2+ab & -ab endvmatrix= (ab+bc+ca)^3$if $p+q+r=0$ Find the value of the DeterminantWhat is the rank of a matrix when the difference between its rows is same?Determinant of beginvmatrix -2a &a+b &a+c \ b+a& -2b &b+c \ c+a&c+b & -2c endvmatrixDefined row operations for making a matrix ,containing varibel coefficients, into RREF
$begingroup$
Consider the $3times 3$ matrix
$$
Aequiv beginpmatrix
mu_1-mu_1'& mu_1-mu_1'-c & mu_1-mu_1'-c-d\
mu_1+a-mu_1'& mu_1+a-mu_1'-c & mu_1+a-mu_1'-c-d\
mu_1+a+b-mu_1'& mu_1+a+b-mu_1'-c & mu_1+a+b-mu_1'-c-d\
endpmatrix
$$
where $mu_1, mu_1', a, b, c, d$ are real numbers and $a,b,c,d$ are strictly positive.
Notice that
$$
R_2=R_1+a\
R_3=R_1+a+b\
C_2=C_1-c\
C_3=C_1-c-d
$$
where $R_1,R_2,R_3,C_1,C_2,C_3$ are respectively the three columns and the 3 rows of $A$.
Claim that I have already shown: Assume
(1) $aneq b,cneq d$
(2) $mu_1,mu_1',a,b,c,d$ are such that, when ordering the elements of $A$ from smallest to largest, the obtained ordered sequence of points is symmetric around zero.
E.g., suppose that $mu_1,mu_1',a,b,c,d$ are such that $A$ contains only $3$ distinct elements which we denote by $eta_1<eta_2<eta_3$. Then, assumption (2) states that
$
eta_1=-eta_3$ and $
eta_2=0$.
Then $mu_1=mu_1', a=c, b=d$. In other words, $C_1=-R_1$,$C_2=-R_2$, $C_3=-R_3$.
Question: can we rewrite $(1)$ [or state sufficient conditions for $(1)$] in a way that involves some "well known" operations on the matrix $A$? E.g., as a sort of rank condition? Or as a sort of determinant condition? Or as a sort of linear independence condition?
More details (not sure they are needed)
I have developed the proof of my claim above in a "very constructive" way which somehow prevents me to understand if there is any " mathematically deeper" meaning of assumption (1) that can be stated in terms of the properties of the matrix $A$.
Let me report here a summary of the proof of my claim above.
1) First of all notice that we can establish a partial order of the elements of $A$, i.e.
$$
Aequiv beginpmatrix
mu_1-mu_1'&<& mu_1-mu_1'-c &<& mu_1-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a-mu_1'&<& mu_1+a-mu_1'-c &<& mu_1+a-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a+b-mu_1'&<& mu_1+a+b-mu_1'-c &<& mu_1+a+b-mu_1'-c-d\
endpmatrix
$$
2) Let $eta_1<eta_2<...<eta_m$ denote the ordered distinct elements of $A$, where $3leq mleq 9$ and
$$
begincases
eta_1equiv mu_1-mu_1'-c-d\
eta_mequiv mu_1+a+b-mu_1'\
endcases
$$
Under assumption (2), $eta_1=-eta_m$, that is
$$
(diamond)hspace1cmmu_1-mu_1'=fracc+d-a-b2
$$
3) The candidates for $eta_2$ are
$$
begincases
mu_1-mu_1'-c\
mu_1+a-mu_1'-c-d\
endcases
$$
and the candidates for $eta_m-1$ are
$$
begincases
mu_1+a-mu_1'\
mu_1+a+b-mu_1'-c
endcases
$$
We know that
$$
small
beginaligned
eta_1<minmu_1-mu_1'-c,
& mu_1+a-mu_1'-c-d leq maxmu_1-mu_1'-c,
mu_1+a-mu_1'-c-d \
& leq
minmu_1+a-mu_1',mu_1+a+b-mu_1'-c leq maxmu_1+a-mu_1',mu_1+a+b-mu_1'-c<eta_m
endaligned
$$
We can show that, under assumption (2) and using $(diamond)$, this implies
$$
mina,d=minb,c
$$
4) Under assumption (1), $mina,d=minb,c$ reduces to two cases
$$
textcase i): b=d<a,d<c
$$
and
$$
textcase ii): a=c<b,c<d
$$
5)
We find $eta_1,...,eta_m$ under case i) and show that assumption (2) implies $mu_1=mu_1'$ and $a=c$.
6)
We find $eta_1,...,eta_m$ under case ii) and show that assumption (2) implies $mu_1=mu_1'$ and $b=d$.
matrices determinant matrix-calculus matrix-decomposition matrix-rank
asked Mar 21 at 14:55
STFSTF
461422
$endgroup$
|
show 1 more comment
$begingroup$
Consider the $3times 3$ matrix
$$
Aequiv beginpmatrix
mu_1-mu_1'& mu_1-mu_1'-c & mu_1-mu_1'-c-d\
mu_1+a-mu_1'& mu_1+a-mu_1'-c & mu_1+a-mu_1'-c-d\
mu_1+a+b-mu_1'& mu_1+a+b-mu_1'-c & mu_1+a+b-mu_1'-c-d\
endpmatrix
$$
where $mu_1, mu_1', a, b, c, d$ are real numbers and $a,b,c,d$ are strictly positive.
Notice that
$$
R_2=R_1+a\
R_3=R_1+a+b\
C_2=C_1-c\
C_3=C_1-c-d
$$
where $R_1,R_2,R_3,C_1,C_2,C_3$ are respectively the three columns and the 3 rows of $A$.
Claim that I have already shown: Assume
(1) $aneq b,cneq d$
(2) $mu_1,mu_1',a,b,c,d$ are such that, when ordering the elements of $A$ from smallest to largest, the obtained ordered sequence of points is symmetric around zero.
E.g., suppose that $mu_1,mu_1',a,b,c,d$ are such that $A$ contains only $3$ distinct elements which we denote by $eta_1<eta_2<eta_3$. Then, assumption (2) states that
$
eta_1=-eta_3$ and $
eta_2=0$.
Then $mu_1=mu_1', a=c, b=d$. In other words, $C_1=-R_1$,$C_2=-R_2$, $C_3=-R_3$.
Question: can we rewrite $(1)$ [or state sufficient conditions for $(1)$] in a way that involves some "well known" operations on the matrix $A$? E.g., as a sort of rank condition? Or as a sort of determinant condition? Or as a sort of linear independence condition?
More details (not sure they are needed)
I have developed the proof of my claim above in a "very constructive" way which somehow prevents me to understand if there is any " mathematically deeper" meaning of assumption (1) that can be stated in terms of the properties of the matrix $A$.
Let me report here a summary of the proof of my claim above.
1) First of all notice that we can establish a partial order of the elements of $A$, i.e.
$$
Aequiv beginpmatrix
mu_1-mu_1'&<& mu_1-mu_1'-c &<& mu_1-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a-mu_1'&<& mu_1+a-mu_1'-c &<& mu_1+a-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a+b-mu_1'&<& mu_1+a+b-mu_1'-c &<& mu_1+a+b-mu_1'-c-d\
endpmatrix
$$
2) Let $eta_1<eta_2<...<eta_m$ denote the ordered distinct elements of $A$, where $3leq mleq 9$ and
$$
begincases
eta_1equiv mu_1-mu_1'-c-d\
eta_mequiv mu_1+a+b-mu_1'\
endcases
$$
Under assumption (2), $eta_1=-eta_m$, that is
$$
(diamond)hspace1cmmu_1-mu_1'=fracc+d-a-b2
$$
3) The candidates for $eta_2$ are
$$
begincases
mu_1-mu_1'-c\
mu_1+a-mu_1'-c-d\
endcases
$$
and the candidates for $eta_m-1$ are
$$
begincases
mu_1+a-mu_1'\
mu_1+a+b-mu_1'-c
endcases
$$
We know that
$$
small
beginaligned
eta_1<minmu_1-mu_1'-c,
& mu_1+a-mu_1'-c-d leq maxmu_1-mu_1'-c,
mu_1+a-mu_1'-c-d \
& leq
minmu_1+a-mu_1',mu_1+a+b-mu_1'-c leq maxmu_1+a-mu_1',mu_1+a+b-mu_1'-c<eta_m
endaligned
$$
We can show that, under assumption (2) and using $(diamond)$, this implies
$$
mina,d=minb,c
$$
4) Under assumption (1), $mina,d=minb,c$ reduces to two cases
$$
textcase i): b=d<a,d<c
$$
and
$$
textcase ii): a=c<b,c<d
$$
5)
We find $eta_1,...,eta_m$ under case i) and show that assumption (2) implies $mu_1=mu_1'$ and $a=c$.
6)
We find $eta_1,...,eta_m$ under case ii) and show that assumption (2) implies $mu_1=mu_1'$ and $b=d$.
matrices determinant matrix-calculus matrix-decomposition matrix-rank
asked Mar 21 at 14:55
STFSTF
461422
$endgroup$
1
$begingroup$
Also, $A = (mu_1-mu_1^prime) beginbmatrix 1&1&1\1&1&1\1&1&1endbmatrix + abeginbmatrix0&0&0\1&1&1\1&1&1endbmatrix +bbeginbmatrix0&0&0\0&0&0\1&1&1endbmatrix -cbeginbmatrix0&1&1\0&1&1\0&1&1endbmatrix -dbeginbmatrix0&0&1\0&0&1\0&0&1endbmatrix$, which will be symmetric iff $a=c$ and $b=d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 18:18
$begingroup$
@MorganRodgers Thanks for asking clarifications. Regarding your first question, I have tried to rewritten assumption (2). Is it more clear now?
$endgroup$
– STF
Mar 21 at 18:21
$begingroup$
@MorganRodgers Regarding your second statement on the symmetry of the matrix $A$: how can it be that $A$ is symmetric when $a=c,b=d$? When $a=c,b=d$, then, for example, $A(2,1)=A(1,2)$ that is $mu_1+c-mu_1'=mu_1-mu_1'-c$ which is impossible given $c>0$.
$endgroup$
– STF
Mar 21 at 18:35
$begingroup$
I apologize, I meant to say $a=-c$, $b=-d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 19:01
$begingroup$
Thanks. In my case it can't be $a=-c,b=-d$ because $a,b,c,d$ are strictly positive. Hence, the matrix $A$ cannot be symmetric.
$endgroup$
– STF
Mar 21 at 19:03
|
show 1 more comment
$begingroup$
Consider the $3times 3$ matrix
$$
Aequiv beginpmatrix
mu_1-mu_1'& mu_1-mu_1'-c & mu_1-mu_1'-c-d\
mu_1+a-mu_1'& mu_1+a-mu_1'-c & mu_1+a-mu_1'-c-d\
mu_1+a+b-mu_1'& mu_1+a+b-mu_1'-c & mu_1+a+b-mu_1'-c-d\
endpmatrix
$$
where $mu_1, mu_1', a, b, c, d$ are real numbers and $a,b,c,d$ are strictly positive.
Notice that
$$
R_2=R_1+a\
R_3=R_1+a+b\
C_2=C_1-c\
C_3=C_1-c-d
$$
where $R_1,R_2,R_3,C_1,C_2,C_3$ are respectively the three columns and the 3 rows of $A$.
Claim that I have already shown: Assume
(1) $aneq b,cneq d$
(2) $mu_1,mu_1',a,b,c,d$ are such that, when ordering the elements of $A$ from smallest to largest, the obtained ordered sequence of points is symmetric around zero.
E.g., suppose that $mu_1,mu_1',a,b,c,d$ are such that $A$ contains only $3$ distinct elements which we denote by $eta_1<eta_2<eta_3$. Then, assumption (2) states that
$
eta_1=-eta_3$ and $
eta_2=0$.
Then $mu_1=mu_1', a=c, b=d$. In other words, $C_1=-R_1$,$C_2=-R_2$, $C_3=-R_3$.
Question: can we rewrite $(1)$ [or state sufficient conditions for $(1)$] in a way that involves some "well known" operations on the matrix $A$? E.g., as a sort of rank condition? Or as a sort of determinant condition? Or as a sort of linear independence condition?
More details (not sure they are needed)
I have developed the proof of my claim above in a "very constructive" way which somehow prevents me to understand if there is any " mathematically deeper" meaning of assumption (1) that can be stated in terms of the properties of the matrix $A$.
Let me report here a summary of the proof of my claim above.
1) First of all notice that we can establish a partial order of the elements of $A$, i.e.
$$
Aequiv beginpmatrix
mu_1-mu_1'&<& mu_1-mu_1'-c &<& mu_1-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a-mu_1'&<& mu_1+a-mu_1'-c &<& mu_1+a-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a+b-mu_1'&<& mu_1+a+b-mu_1'-c &<& mu_1+a+b-mu_1'-c-d\
endpmatrix
$$
2) Let $eta_1<eta_2<...<eta_m$ denote the ordered distinct elements of $A$, where $3leq mleq 9$ and
$$
begincases
eta_1equiv mu_1-mu_1'-c-d\
eta_mequiv mu_1+a+b-mu_1'\
endcases
$$
Under assumption (2), $eta_1=-eta_m$, that is
$$
(diamond)hspace1cmmu_1-mu_1'=fracc+d-a-b2
$$
3) The candidates for $eta_2$ are
$$
begincases
mu_1-mu_1'-c\
mu_1+a-mu_1'-c-d\
endcases
$$
and the candidates for $eta_m-1$ are
$$
begincases
mu_1+a-mu_1'\
mu_1+a+b-mu_1'-c
endcases
$$
We know that
$$
small
beginaligned
eta_1<minmu_1-mu_1'-c,
& mu_1+a-mu_1'-c-d leq maxmu_1-mu_1'-c,
mu_1+a-mu_1'-c-d \
& leq
minmu_1+a-mu_1',mu_1+a+b-mu_1'-c leq maxmu_1+a-mu_1',mu_1+a+b-mu_1'-c<eta_m
endaligned
$$
We can show that, under assumption (2) and using $(diamond)$, this implies
$$
mina,d=minb,c
$$
4) Under assumption (1), $mina,d=minb,c$ reduces to two cases
$$
textcase i): b=d<a,d<c
$$
and
$$
textcase ii): a=c<b,c<d
$$
5)
We find $eta_1,...,eta_m$ under case i) and show that assumption (2) implies $mu_1=mu_1'$ and $a=c$.
6)
We find $eta_1,...,eta_m$ under case ii) and show that assumption (2) implies $mu_1=mu_1'$ and $b=d$.
matrices determinant matrix-calculus matrix-decomposition matrix-rank
asked Mar 21 at 14:55
STFSTF
461422
$endgroup$
Consider the $3times 3$ matrix
$$
Aequiv beginpmatrix
mu_1-mu_1'& mu_1-mu_1'-c & mu_1-mu_1'-c-d\
mu_1+a-mu_1'& mu_1+a-mu_1'-c & mu_1+a-mu_1'-c-d\
mu_1+a+b-mu_1'& mu_1+a+b-mu_1'-c & mu_1+a+b-mu_1'-c-d\
endpmatrix
$$
where $mu_1, mu_1', a, b, c, d$ are real numbers and $a,b,c,d$ are strictly positive.
Notice that
$$
R_2=R_1+a\
R_3=R_1+a+b\
C_2=C_1-c\
C_3=C_1-c-d
$$
where $R_1,R_2,R_3,C_1,C_2,C_3$ are respectively the three columns and the 3 rows of $A$.
Claim that I have already shown: Assume
(1) $aneq b,cneq d$
(2) $mu_1,mu_1',a,b,c,d$ are such that, when ordering the elements of $A$ from smallest to largest, the obtained ordered sequence of points is symmetric around zero.
E.g., suppose that $mu_1,mu_1',a,b,c,d$ are such that $A$ contains only $3$ distinct elements which we denote by $eta_1<eta_2<eta_3$. Then, assumption (2) states that
$
eta_1=-eta_3$ and $
eta_2=0$.
Then $mu_1=mu_1', a=c, b=d$. In other words, $C_1=-R_1$,$C_2=-R_2$, $C_3=-R_3$.
Question: can we rewrite $(1)$ [or state sufficient conditions for $(1)$] in a way that involves some "well known" operations on the matrix $A$? E.g., as a sort of rank condition? Or as a sort of determinant condition? Or as a sort of linear independence condition?
More details (not sure they are needed)
I have developed the proof of my claim above in a "very constructive" way which somehow prevents me to understand if there is any " mathematically deeper" meaning of assumption (1) that can be stated in terms of the properties of the matrix $A$.
Let me report here a summary of the proof of my claim above.
1) First of all notice that we can establish a partial order of the elements of $A$, i.e.
$$
Aequiv beginpmatrix
mu_1-mu_1'&<& mu_1-mu_1'-c &<& mu_1-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a-mu_1'&<& mu_1+a-mu_1'-c &<& mu_1+a-mu_1'-c-d\
wedge && wedge && wedge\
mu_1+a+b-mu_1'&<& mu_1+a+b-mu_1'-c &<& mu_1+a+b-mu_1'-c-d\
endpmatrix
$$
2) Let $eta_1<eta_2<...<eta_m$ denote the ordered distinct elements of $A$, where $3leq mleq 9$ and
$$
begincases
eta_1equiv mu_1-mu_1'-c-d\
eta_mequiv mu_1+a+b-mu_1'\
endcases
$$
Under assumption (2), $eta_1=-eta_m$, that is
$$
(diamond)hspace1cmmu_1-mu_1'=fracc+d-a-b2
$$
3) The candidates for $eta_2$ are
$$
begincases
mu_1-mu_1'-c\
mu_1+a-mu_1'-c-d\
endcases
$$
and the candidates for $eta_m-1$ are
$$
begincases
mu_1+a-mu_1'\
mu_1+a+b-mu_1'-c
endcases
$$
We know that
$$
small
beginaligned
eta_1<minmu_1-mu_1'-c,
& mu_1+a-mu_1'-c-d leq maxmu_1-mu_1'-c,
mu_1+a-mu_1'-c-d \
& leq
minmu_1+a-mu_1',mu_1+a+b-mu_1'-c leq maxmu_1+a-mu_1',mu_1+a+b-mu_1'-c<eta_m
endaligned
$$
We can show that, under assumption (2) and using $(diamond)$, this implies
$$
mina,d=minb,c
$$
4) Under assumption (1), $mina,d=minb,c$ reduces to two cases
$$
textcase i): b=d<a,d<c
$$
and
$$
textcase ii): a=c<b,c<d
$$
5)
We find $eta_1,...,eta_m$ under case i) and show that assumption (2) implies $mu_1=mu_1'$ and $a=c$.
6)
We find $eta_1,...,eta_m$ under case ii) and show that assumption (2) implies $mu_1=mu_1'$ and $b=d$.
matrices determinant matrix-calculus matrix-decomposition matrix-rank
matrices determinant matrix-calculus matrix-decomposition matrix-rank
asked Mar 21 at 14:55
STFSTF
461422
asked Mar 21 at 14:55
STFSTF
461422
edited Mar 21 at 18:44
STF
asked Mar 21 at 14:55
STFSTF
461422
asked Mar 21 at 14:55
STFSTF
461422
asked Mar 21 at 14:55
STFSTF
461422
461422
1
$begingroup$
Also, $A = (mu_1-mu_1^prime) beginbmatrix 1&1&1\1&1&1\1&1&1endbmatrix + abeginbmatrix0&0&0\1&1&1\1&1&1endbmatrix +bbeginbmatrix0&0&0\0&0&0\1&1&1endbmatrix -cbeginbmatrix0&1&1\0&1&1\0&1&1endbmatrix -dbeginbmatrix0&0&1\0&0&1\0&0&1endbmatrix$, which will be symmetric iff $a=c$ and $b=d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 18:18
$begingroup$
@MorganRodgers Thanks for asking clarifications. Regarding your first question, I have tried to rewritten assumption (2). Is it more clear now?
$endgroup$
– STF
Mar 21 at 18:21
$begingroup$
@MorganRodgers Regarding your second statement on the symmetry of the matrix $A$: how can it be that $A$ is symmetric when $a=c,b=d$? When $a=c,b=d$, then, for example, $A(2,1)=A(1,2)$ that is $mu_1+c-mu_1'=mu_1-mu_1'-c$ which is impossible given $c>0$.
$endgroup$
– STF
Mar 21 at 18:35
$begingroup$
I apologize, I meant to say $a=-c$, $b=-d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 19:01
$begingroup$
Thanks. In my case it can't be $a=-c,b=-d$ because $a,b,c,d$ are strictly positive. Hence, the matrix $A$ cannot be symmetric.
$endgroup$
– STF
Mar 21 at 19:03
|
show 1 more comment
1
$begingroup$
Also, $A = (mu_1-mu_1^prime) beginbmatrix 1&1&1\1&1&1\1&1&1endbmatrix + abeginbmatrix0&0&0\1&1&1\1&1&1endbmatrix +bbeginbmatrix0&0&0\0&0&0\1&1&1endbmatrix -cbeginbmatrix0&1&1\0&1&1\0&1&1endbmatrix -dbeginbmatrix0&0&1\0&0&1\0&0&1endbmatrix$, which will be symmetric iff $a=c$ and $b=d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 18:18
$begingroup$
@MorganRodgers Thanks for asking clarifications. Regarding your first question, I have tried to rewritten assumption (2). Is it more clear now?
$endgroup$
– STF
Mar 21 at 18:21
$begingroup$
@MorganRodgers Regarding your second statement on the symmetry of the matrix $A$: how can it be that $A$ is symmetric when $a=c,b=d$? When $a=c,b=d$, then, for example, $A(2,1)=A(1,2)$ that is $mu_1+c-mu_1'=mu_1-mu_1'-c$ which is impossible given $c>0$.
$endgroup$
– STF
Mar 21 at 18:35
$begingroup$
I apologize, I meant to say $a=-c$, $b=-d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 19:01
$begingroup$
Thanks. In my case it can't be $a=-c,b=-d$ because $a,b,c,d$ are strictly positive. Hence, the matrix $A$ cannot be symmetric.
$endgroup$
– STF
Mar 21 at 19:03
1
1
$begingroup$
Also, $A = (mu_1-mu_1^prime) beginbmatrix 1&1&1\1&1&1\1&1&1endbmatrix + abeginbmatrix0&0&0\1&1&1\1&1&1endbmatrix +bbeginbmatrix0&0&0\0&0&0\1&1&1endbmatrix -cbeginbmatrix0&1&1\0&1&1\0&1&1endbmatrix -dbeginbmatrix0&0&1\0&0&1\0&0&1endbmatrix$, which will be symmetric iff $a=c$ and $b=d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 18:18
$begingroup$
Also, $A = (mu_1-mu_1^prime) beginbmatrix 1&1&1\1&1&1\1&1&1endbmatrix + abeginbmatrix0&0&0\1&1&1\1&1&1endbmatrix +bbeginbmatrix0&0&0\0&0&0\1&1&1endbmatrix -cbeginbmatrix0&1&1\0&1&1\0&1&1endbmatrix -dbeginbmatrix0&0&1\0&0&1\0&0&1endbmatrix$, which will be symmetric iff $a=c$ and $b=d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 18:18
$begingroup$
@MorganRodgers Thanks for asking clarifications. Regarding your first question, I have tried to rewritten assumption (2). Is it more clear now?
$endgroup$
– STF
Mar 21 at 18:21
$begingroup$
@MorganRodgers Thanks for asking clarifications. Regarding your first question, I have tried to rewritten assumption (2). Is it more clear now?
$endgroup$
– STF
Mar 21 at 18:21
$begingroup$
@MorganRodgers Regarding your second statement on the symmetry of the matrix $A$: how can it be that $A$ is symmetric when $a=c,b=d$? When $a=c,b=d$, then, for example, $A(2,1)=A(1,2)$ that is $mu_1+c-mu_1'=mu_1-mu_1'-c$ which is impossible given $c>0$.
$endgroup$
– STF
Mar 21 at 18:35
$begingroup$
@MorganRodgers Regarding your second statement on the symmetry of the matrix $A$: how can it be that $A$ is symmetric when $a=c,b=d$? When $a=c,b=d$, then, for example, $A(2,1)=A(1,2)$ that is $mu_1+c-mu_1'=mu_1-mu_1'-c$ which is impossible given $c>0$.
$endgroup$
– STF
Mar 21 at 18:35
$begingroup$
I apologize, I meant to say $a=-c$, $b=-d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 19:01
$begingroup$
I apologize, I meant to say $a=-c$, $b=-d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 19:01
$begingroup$
Thanks. In my case it can't be $a=-c,b=-d$ because $a,b,c,d$ are strictly positive. Hence, the matrix $A$ cannot be symmetric.
$endgroup$
– STF
Mar 21 at 19:03
$begingroup$
Thanks. In my case it can't be $a=-c,b=-d$ because $a,b,c,d$ are strictly positive. Hence, the matrix $A$ cannot be symmetric.
$endgroup$
– STF
Mar 21 at 19:03
|
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1
$begingroup$
Also, $A = (mu_1-mu_1^prime) beginbmatrix 1&1&1\1&1&1\1&1&1endbmatrix + abeginbmatrix0&0&0\1&1&1\1&1&1endbmatrix +bbeginbmatrix0&0&0\0&0&0\1&1&1endbmatrix -cbeginbmatrix0&1&1\0&1&1\0&1&1endbmatrix -dbeginbmatrix0&0&1\0&0&1\0&0&1endbmatrix$, which will be symmetric iff $a=c$ and $b=d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 18:18
$begingroup$
@MorganRodgers Thanks for asking clarifications. Regarding your first question, I have tried to rewritten assumption (2). Is it more clear now?
$endgroup$
– STF
Mar 21 at 18:21
$begingroup$
@MorganRodgers Regarding your second statement on the symmetry of the matrix $A$: how can it be that $A$ is symmetric when $a=c,b=d$? When $a=c,b=d$, then, for example, $A(2,1)=A(1,2)$ that is $mu_1+c-mu_1'=mu_1-mu_1'-c$ which is impossible given $c>0$.
$endgroup$
– STF
Mar 21 at 18:35
$begingroup$
I apologize, I meant to say $a=-c$, $b=-d$.
$endgroup$
– Morgan Rodgers
Mar 21 at 19:01
$begingroup$
Thanks. In my case it can't be $a=-c,b=-d$ because $a,b,c,d$ are strictly positive. Hence, the matrix $A$ cannot be symmetric.
$endgroup$
– STF
Mar 21 at 19:03