Mixing
How to prove this matrix is rank n?
$begingroup$
Let $M = begin{pmatrix} p_0 (1- p_0) & -p_0 p_1 &ldots & p_0 p_n\
-p_1 p_0 & p_1 (1-p_1) & ldots & p_1 p_n\
vdots & & &vdots\
-p_np_0 &ldots&&p_n(1-p_n)end{pmatrix}$ where $sum_i p_i = 1$ and $p_i >0$ for all i. How to show this matrix has rank n?
For clarity, $M_{i,j} = p_i(delta_{ij} - p_j)$ and $delta_{ij} = 1$ if $i = j$ and 0 otherwise, where $sum_i p_i = 1$ and $p_i >0$.
I tried random values of ${p_0, ldots, p_n}$ and it consistently returns n, so hoping for the rank be n.
linear-algebra matrices matrix-rank
asked Jan 31 at 6:06
listenerlistener
205
$endgroup$
add a comment |
$begingroup$
Let $M = begin{pmatrix} p_0 (1- p_0) & -p_0 p_1 &ldots & p_0 p_n\
-p_1 p_0 & p_1 (1-p_1) & ldots & p_1 p_n\
vdots & & &vdots\
-p_np_0 &ldots&&p_n(1-p_n)end{pmatrix}$ where $sum_i p_i = 1$ and $p_i >0$ for all i. How to show this matrix has rank n?
For clarity, $M_{i,j} = p_i(delta_{ij} - p_j)$ and $delta_{ij} = 1$ if $i = j$ and 0 otherwise, where $sum_i p_i = 1$ and $p_i >0$.
I tried random values of ${p_0, ldots, p_n}$ and it consistently returns n, so hoping for the rank be n.
linear-algebra matrices matrix-rank
asked Jan 31 at 6:06
listenerlistener
205
$endgroup$
2
$begingroup$
The matrix as you have written it is $(n+1)times(n+1)$. If $n=1$ it's a $2times2$ matrix with rank $1$. I think you mean to conjecture that the matrix has rank $n$.
$endgroup$
– David
Jan 31 at 6:15
$begingroup$
yes, I will update the question
$endgroup$
– listener
Jan 31 at 15:13
add a comment |
$begingroup$
Let $M = begin{pmatrix} p_0 (1- p_0) & -p_0 p_1 &ldots & p_0 p_n\
-p_1 p_0 & p_1 (1-p_1) & ldots & p_1 p_n\
vdots & & &vdots\
-p_np_0 &ldots&&p_n(1-p_n)end{pmatrix}$ where $sum_i p_i = 1$ and $p_i >0$ for all i. How to show this matrix has rank n?
For clarity, $M_{i,j} = p_i(delta_{ij} - p_j)$ and $delta_{ij} = 1$ if $i = j$ and 0 otherwise, where $sum_i p_i = 1$ and $p_i >0$.
I tried random values of ${p_0, ldots, p_n}$ and it consistently returns n, so hoping for the rank be n.
linear-algebra matrices matrix-rank
asked Jan 31 at 6:06
listenerlistener
205
$endgroup$
Let $M = begin{pmatrix} p_0 (1- p_0) & -p_0 p_1 &ldots & p_0 p_n\
-p_1 p_0 & p_1 (1-p_1) & ldots & p_1 p_n\
vdots & & &vdots\
-p_np_0 &ldots&&p_n(1-p_n)end{pmatrix}$ where $sum_i p_i = 1$ and $p_i >0$ for all i. How to show this matrix has rank n?
For clarity, $M_{i,j} = p_i(delta_{ij} - p_j)$ and $delta_{ij} = 1$ if $i = j$ and 0 otherwise, where $sum_i p_i = 1$ and $p_i >0$.
I tried random values of ${p_0, ldots, p_n}$ and it consistently returns n, so hoping for the rank be n.
linear-algebra matrices matrix-rank
linear-algebra matrices matrix-rank
asked Jan 31 at 6:06
listenerlistener
205
asked Jan 31 at 6:06
listenerlistener
205
edited Jan 31 at 15:14
listener
asked Jan 31 at 6:06
listenerlistener
205
asked Jan 31 at 6:06
listenerlistener
205
asked Jan 31 at 6:06
listenerlistener
205
205
2
$begingroup$
The matrix as you have written it is $(n+1)times(n+1)$. If $n=1$ it's a $2times2$ matrix with rank $1$. I think you mean to conjecture that the matrix has rank $n$.
$endgroup$
– David
Jan 31 at 6:15
$begingroup$
yes, I will update the question
$endgroup$
– listener
Jan 31 at 15:13
add a comment |
2
$begingroup$
The matrix as you have written it is $(n+1)times(n+1)$. If $n=1$ it's a $2times2$ matrix with rank $1$. I think you mean to conjecture that the matrix has rank $n$.
$endgroup$
– David
Jan 31 at 6:15
$begingroup$
yes, I will update the question
$endgroup$
– listener
Jan 31 at 15:13
2
2
$begingroup$
The matrix as you have written it is $(n+1)times(n+1)$. If $n=1$ it's a $2times2$ matrix with rank $1$. I think you mean to conjecture that the matrix has rank $n$.
$endgroup$
– David
Jan 31 at 6:15
$begingroup$
The matrix as you have written it is $(n+1)times(n+1)$. If $n=1$ it's a $2times2$ matrix with rank $1$. I think you mean to conjecture that the matrix has rank $n$.
$endgroup$
– David
Jan 31 at 6:15
$begingroup$
yes, I will update the question
$endgroup$
– listener
Jan 31 at 15:13
$begingroup$
yes, I will update the question
$endgroup$
– listener
Jan 31 at 15:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your matrix is $$M=D-uu^top$$
, where $D=text{diag}(p_0,p_1,ldots,p_n)$ and $u=(p_0,p_1,ldots,p_n)^top$
Consequently, $$operatorname{rank}(M)ge |text{rank}(D)-operatorname{rank}(uu^top)|=n+1-1=n$$
The matrix $M$ of order $(n+1)times(n+1)$ is related to the dispersion matrix of a singular multinomial distribution ('singular' because of the restriction $sumlimits_{k=0}^n p_k=1$). Hence rank of $M$ is at most $n$.
This is also seen from the fact that $$det M=left(prod_{k=0}^n p_kright)left(1-sum_{k=0}^n p_kright)=0$$
So, $operatorname{rank}(M)=n$.
answered Jan 31 at 6:38
StubbornAtomStubbornAtom
6,29831440
$endgroup$
1
$begingroup$
Thanks for the solution. Can you elaborate on $rank(M) geq n$? I agree rank(D) = n+1 and $rank(uu^T) =1$ but how is $rank(M) geq | rank(D) - rank(u u^T)|$?
$endgroup$
– listener
Jan 31 at 15:24
$begingroup$
@listener It follows from $$text{rank}(color{green}{A-B}+color{blue}B)le text{rank}(color{green}{A-B})+text{rank}(color{blue}B)$$ for any two matrices $A$ and $B$ having the same order.
$endgroup$
– StubbornAtom
Jan 31 at 15:28
$begingroup$
is there a citation for this theorem please or some name how I can find this? Thanks
$endgroup$
– listener
Jan 31 at 15:34
$begingroup$
@listener math.stackexchange.com/questions/851596/….
$endgroup$
– StubbornAtom
Jan 31 at 15:35
$begingroup$
Awesome, Thanks! Also the matrix determinant lemma works only for $A + uu^T$, you seem to be using it for $A-uu^T$. How do you deduce that?
$endgroup$
– listener
Jan 31 at 15:51
|
show 2 more comments
$begingroup$
For a start: if you add every other row to the first row you get a zero row because
$$p_0(1-p_0)-p_1p_0-cdots-p_np_0=p_0-p_0(p_0+p_1+cdots+p_n)=0$$
in the first column, and similarly for all other columns. So the rank of the $(n+1)times(n+1)$ matrix is at most $n$.
Sorry but I gotta go now. Hope somebody else can finish the problem by showing that the rank is not less than $n$.
answered Jan 31 at 6:18
DavidDavid
69.7k668131
$endgroup$
$begingroup$
To complete David answer, once we know the rank is $<n+1$, it suffices to find a $ntimes n$ sub-matrix which has a non-zero determinant.
$endgroup$
– Clément Guérin
Jan 31 at 6:21
add a comment |
Your Answer
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2 Answers
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2 Answers
2
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$begingroup$
Your matrix is $$M=D-uu^top$$
, where $D=text{diag}(p_0,p_1,ldots,p_n)$ and $u=(p_0,p_1,ldots,p_n)^top$
Consequently, $$operatorname{rank}(M)ge |text{rank}(D)-operatorname{rank}(uu^top)|=n+1-1=n$$
The matrix $M$ of order $(n+1)times(n+1)$ is related to the dispersion matrix of a singular multinomial distribution ('singular' because of the restriction $sumlimits_{k=0}^n p_k=1$). Hence rank of $M$ is at most $n$.
This is also seen from the fact that $$det M=left(prod_{k=0}^n p_kright)left(1-sum_{k=0}^n p_kright)=0$$
So, $operatorname{rank}(M)=n$.
answered Jan 31 at 6:38
StubbornAtomStubbornAtom
6,29831440
$endgroup$
1
$begingroup$
Thanks for the solution. Can you elaborate on $rank(M) geq n$? I agree rank(D) = n+1 and $rank(uu^T) =1$ but how is $rank(M) geq | rank(D) - rank(u u^T)|$?
$endgroup$
– listener
Jan 31 at 15:24
$begingroup$
@listener It follows from $$text{rank}(color{green}{A-B}+color{blue}B)le text{rank}(color{green}{A-B})+text{rank}(color{blue}B)$$ for any two matrices $A$ and $B$ having the same order.
$endgroup$
– StubbornAtom
Jan 31 at 15:28
$begingroup$
is there a citation for this theorem please or some name how I can find this? Thanks
$endgroup$
– listener
Jan 31 at 15:34
$begingroup$
@listener math.stackexchange.com/questions/851596/….
$endgroup$
– StubbornAtom
Jan 31 at 15:35
$begingroup$
Awesome, Thanks! Also the matrix determinant lemma works only for $A + uu^T$, you seem to be using it for $A-uu^T$. How do you deduce that?
$endgroup$
– listener
Jan 31 at 15:51
|
show 2 more comments
$begingroup$
Your matrix is $$M=D-uu^top$$
, where $D=text{diag}(p_0,p_1,ldots,p_n)$ and $u=(p_0,p_1,ldots,p_n)^top$
Consequently, $$operatorname{rank}(M)ge |text{rank}(D)-operatorname{rank}(uu^top)|=n+1-1=n$$
The matrix $M$ of order $(n+1)times(n+1)$ is related to the dispersion matrix of a singular multinomial distribution ('singular' because of the restriction $sumlimits_{k=0}^n p_k=1$). Hence rank of $M$ is at most $n$.
This is also seen from the fact that $$det M=left(prod_{k=0}^n p_kright)left(1-sum_{k=0}^n p_kright)=0$$
So, $operatorname{rank}(M)=n$.
answered Jan 31 at 6:38
StubbornAtomStubbornAtom
6,29831440
$endgroup$
1
$begingroup$
Thanks for the solution. Can you elaborate on $rank(M) geq n$? I agree rank(D) = n+1 and $rank(uu^T) =1$ but how is $rank(M) geq | rank(D) - rank(u u^T)|$?
$endgroup$
– listener
Jan 31 at 15:24
$begingroup$
@listener It follows from $$text{rank}(color{green}{A-B}+color{blue}B)le text{rank}(color{green}{A-B})+text{rank}(color{blue}B)$$ for any two matrices $A$ and $B$ having the same order.
$endgroup$
– StubbornAtom
Jan 31 at 15:28
$begingroup$
is there a citation for this theorem please or some name how I can find this? Thanks
$endgroup$
– listener
Jan 31 at 15:34
$begingroup$
@listener math.stackexchange.com/questions/851596/….
$endgroup$
– StubbornAtom
Jan 31 at 15:35
$begingroup$
Awesome, Thanks! Also the matrix determinant lemma works only for $A + uu^T$, you seem to be using it for $A-uu^T$. How do you deduce that?
$endgroup$
– listener
Jan 31 at 15:51
|
show 2 more comments
$begingroup$
Your matrix is $$M=D-uu^top$$
, where $D=text{diag}(p_0,p_1,ldots,p_n)$ and $u=(p_0,p_1,ldots,p_n)^top$
Consequently, $$operatorname{rank}(M)ge |text{rank}(D)-operatorname{rank}(uu^top)|=n+1-1=n$$
The matrix $M$ of order $(n+1)times(n+1)$ is related to the dispersion matrix of a singular multinomial distribution ('singular' because of the restriction $sumlimits_{k=0}^n p_k=1$). Hence rank of $M$ is at most $n$.
This is also seen from the fact that $$det M=left(prod_{k=0}^n p_kright)left(1-sum_{k=0}^n p_kright)=0$$
So, $operatorname{rank}(M)=n$.
answered Jan 31 at 6:38
StubbornAtomStubbornAtom
6,29831440
$endgroup$
Your matrix is $$M=D-uu^top$$
, where $D=text{diag}(p_0,p_1,ldots,p_n)$ and $u=(p_0,p_1,ldots,p_n)^top$
Consequently, $$operatorname{rank}(M)ge |text{rank}(D)-operatorname{rank}(uu^top)|=n+1-1=n$$
The matrix $M$ of order $(n+1)times(n+1)$ is related to the dispersion matrix of a singular multinomial distribution ('singular' because of the restriction $sumlimits_{k=0}^n p_k=1$). Hence rank of $M$ is at most $n$.
This is also seen from the fact that $$det M=left(prod_{k=0}^n p_kright)left(1-sum_{k=0}^n p_kright)=0$$
So, $operatorname{rank}(M)=n$.
answered Jan 31 at 6:38
StubbornAtomStubbornAtom
6,29831440
edited Jan 31 at 6:53
answered Jan 31 at 6:38
StubbornAtomStubbornAtom
6,29831440
answered Jan 31 at 6:38
StubbornAtomStubbornAtom
6,29831440
answered Jan 31 at 6:38
StubbornAtomStubbornAtom
6,29831440
6,29831440
1
$begingroup$
Thanks for the solution. Can you elaborate on $rank(M) geq n$? I agree rank(D) = n+1 and $rank(uu^T) =1$ but how is $rank(M) geq | rank(D) - rank(u u^T)|$?
$endgroup$
– listener
Jan 31 at 15:24
$begingroup$
@listener It follows from $$text{rank}(color{green}{A-B}+color{blue}B)le text{rank}(color{green}{A-B})+text{rank}(color{blue}B)$$ for any two matrices $A$ and $B$ having the same order.
$endgroup$
– StubbornAtom
Jan 31 at 15:28
$begingroup$
is there a citation for this theorem please or some name how I can find this? Thanks
$endgroup$
– listener
Jan 31 at 15:34
$begingroup$
@listener math.stackexchange.com/questions/851596/….
$endgroup$
– StubbornAtom
Jan 31 at 15:35
$begingroup$
Awesome, Thanks! Also the matrix determinant lemma works only for $A + uu^T$, you seem to be using it for $A-uu^T$. How do you deduce that?
$endgroup$
– listener
Jan 31 at 15:51
|
show 2 more comments
1
$begingroup$
Thanks for the solution. Can you elaborate on $rank(M) geq n$? I agree rank(D) = n+1 and $rank(uu^T) =1$ but how is $rank(M) geq | rank(D) - rank(u u^T)|$?
$endgroup$
– listener
Jan 31 at 15:24
$begingroup$
@listener It follows from $$text{rank}(color{green}{A-B}+color{blue}B)le text{rank}(color{green}{A-B})+text{rank}(color{blue}B)$$ for any two matrices $A$ and $B$ having the same order.
$endgroup$
– StubbornAtom
Jan 31 at 15:28
$begingroup$
is there a citation for this theorem please or some name how I can find this? Thanks
$endgroup$
– listener
Jan 31 at 15:34
$begingroup$
@listener math.stackexchange.com/questions/851596/….
$endgroup$
– StubbornAtom
Jan 31 at 15:35
$begingroup$
Awesome, Thanks! Also the matrix determinant lemma works only for $A + uu^T$, you seem to be using it for $A-uu^T$. How do you deduce that?
$endgroup$
– listener
Jan 31 at 15:51
1
1
$begingroup$
Thanks for the solution. Can you elaborate on $rank(M) geq n$? I agree rank(D) = n+1 and $rank(uu^T) =1$ but how is $rank(M) geq | rank(D) - rank(u u^T)|$?
$endgroup$
– listener
Jan 31 at 15:24
$begingroup$
Thanks for the solution. Can you elaborate on $rank(M) geq n$? I agree rank(D) = n+1 and $rank(uu^T) =1$ but how is $rank(M) geq | rank(D) - rank(u u^T)|$?
$endgroup$
– listener
Jan 31 at 15:24
$begingroup$
@listener It follows from $$text{rank}(color{green}{A-B}+color{blue}B)le text{rank}(color{green}{A-B})+text{rank}(color{blue}B)$$ for any two matrices $A$ and $B$ having the same order.
$endgroup$
– StubbornAtom
Jan 31 at 15:28
$begingroup$
@listener It follows from $$text{rank}(color{green}{A-B}+color{blue}B)le text{rank}(color{green}{A-B})+text{rank}(color{blue}B)$$ for any two matrices $A$ and $B$ having the same order.
$endgroup$
– StubbornAtom
Jan 31 at 15:28
$begingroup$
is there a citation for this theorem please or some name how I can find this? Thanks
$endgroup$
– listener
Jan 31 at 15:34
$begingroup$
is there a citation for this theorem please or some name how I can find this? Thanks
$endgroup$
– listener
Jan 31 at 15:34
$begingroup$
@listener math.stackexchange.com/questions/851596/….
$endgroup$
– StubbornAtom
Jan 31 at 15:35
$begingroup$
@listener math.stackexchange.com/questions/851596/….
$endgroup$
– StubbornAtom
Jan 31 at 15:35
$begingroup$
Awesome, Thanks! Also the matrix determinant lemma works only for $A + uu^T$, you seem to be using it for $A-uu^T$. How do you deduce that?
$endgroup$
– listener
Jan 31 at 15:51
$begingroup$
Awesome, Thanks! Also the matrix determinant lemma works only for $A + uu^T$, you seem to be using it for $A-uu^T$. How do you deduce that?
$endgroup$
– listener
Jan 31 at 15:51
|
show 2 more comments
$begingroup$
For a start: if you add every other row to the first row you get a zero row because
$$p_0(1-p_0)-p_1p_0-cdots-p_np_0=p_0-p_0(p_0+p_1+cdots+p_n)=0$$
in the first column, and similarly for all other columns. So the rank of the $(n+1)times(n+1)$ matrix is at most $n$.
Sorry but I gotta go now. Hope somebody else can finish the problem by showing that the rank is not less than $n$.
answered Jan 31 at 6:18
DavidDavid
69.7k668131
$endgroup$
$begingroup$
To complete David answer, once we know the rank is $<n+1$, it suffices to find a $ntimes n$ sub-matrix which has a non-zero determinant.
$endgroup$
– Clément Guérin
Jan 31 at 6:21
add a comment |
$begingroup$
For a start: if you add every other row to the first row you get a zero row because
$$p_0(1-p_0)-p_1p_0-cdots-p_np_0=p_0-p_0(p_0+p_1+cdots+p_n)=0$$
in the first column, and similarly for all other columns. So the rank of the $(n+1)times(n+1)$ matrix is at most $n$.
Sorry but I gotta go now. Hope somebody else can finish the problem by showing that the rank is not less than $n$.
answered Jan 31 at 6:18
DavidDavid
69.7k668131
$endgroup$
$begingroup$
To complete David answer, once we know the rank is $<n+1$, it suffices to find a $ntimes n$ sub-matrix which has a non-zero determinant.
$endgroup$
– Clément Guérin
Jan 31 at 6:21
add a comment |
$begingroup$
For a start: if you add every other row to the first row you get a zero row because
$$p_0(1-p_0)-p_1p_0-cdots-p_np_0=p_0-p_0(p_0+p_1+cdots+p_n)=0$$
in the first column, and similarly for all other columns. So the rank of the $(n+1)times(n+1)$ matrix is at most $n$.
Sorry but I gotta go now. Hope somebody else can finish the problem by showing that the rank is not less than $n$.
answered Jan 31 at 6:18
DavidDavid
69.7k668131
$endgroup$
For a start: if you add every other row to the first row you get a zero row because
$$p_0(1-p_0)-p_1p_0-cdots-p_np_0=p_0-p_0(p_0+p_1+cdots+p_n)=0$$
in the first column, and similarly for all other columns. So the rank of the $(n+1)times(n+1)$ matrix is at most $n$.
Sorry but I gotta go now. Hope somebody else can finish the problem by showing that the rank is not less than $n$.
answered Jan 31 at 6:18
DavidDavid
69.7k668131
answered Jan 31 at 6:18
DavidDavid
69.7k668131
answered Jan 31 at 6:18
DavidDavid
69.7k668131
answered Jan 31 at 6:18
DavidDavid
69.7k668131
69.7k668131
$begingroup$
To complete David answer, once we know the rank is $<n+1$, it suffices to find a $ntimes n$ sub-matrix which has a non-zero determinant.
$endgroup$
– Clément Guérin
Jan 31 at 6:21
add a comment |
$begingroup$
To complete David answer, once we know the rank is $<n+1$, it suffices to find a $ntimes n$ sub-matrix which has a non-zero determinant.
$endgroup$
– Clément Guérin
Jan 31 at 6:21
$begingroup$
To complete David answer, once we know the rank is $<n+1$, it suffices to find a $ntimes n$ sub-matrix which has a non-zero determinant.
$endgroup$
– Clément Guérin
Jan 31 at 6:21
$begingroup$
To complete David answer, once we know the rank is $<n+1$, it suffices to find a $ntimes n$ sub-matrix which has a non-zero determinant.
$endgroup$
– Clément Guérin
Jan 31 at 6:21
add a comment |
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2
$begingroup$
The matrix as you have written it is $(n+1)times(n+1)$. If $n=1$ it's a $2times2$ matrix with rank $1$. I think you mean to conjecture that the matrix has rank $n$.
$endgroup$
– David
Jan 31 at 6:15
$begingroup$
yes, I will update the question
$endgroup$
– listener
Jan 31 at 15:13