Decomposing a positive semi-definite matrix with all -1,+1 elements
up vote
2
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Claim.$,$ A matrix $,X in {-1,1}^{ntimes n},$ is positive semi-definite if and only if it is of the form $X= xx^{T}$, for some $x in {-1,1 }^n$.
How can I prove this? Proving the 'if' part is easy since $y^Txx^Ty = (x^Ty)^T(x^Ty) geq 0$ for any $y in mathbb{R}^n$. But, the other way around is not that straightforward.
Notice that, $X$ has all ones diagonal since diagonal of a psd matrix cannot have negative elements.
linear-algebra matrices matrix-calculus symmetric-matrices positive-semidefinite
edited Dec 3 at 19:28
Yiorgos S. Smyrlis
62.3k1383162
asked Nov 22 at 18:45
independentvariable
13210
add a comment |
up vote
2
down vote
favorite
Claim.$,$ A matrix $,X in {-1,1}^{ntimes n},$ is positive semi-definite if and only if it is of the form $X= xx^{T}$, for some $x in {-1,1 }^n$.
How can I prove this? Proving the 'if' part is easy since $y^Txx^Ty = (x^Ty)^T(x^Ty) geq 0$ for any $y in mathbb{R}^n$. But, the other way around is not that straightforward.
Notice that, $X$ has all ones diagonal since diagonal of a psd matrix cannot have negative elements.
linear-algebra matrices matrix-calculus symmetric-matrices positive-semidefinite
edited Dec 3 at 19:28
Yiorgos S. Smyrlis
62.3k1383162
asked Nov 22 at 18:45
independentvariable
13210
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Claim.$,$ A matrix $,X in {-1,1}^{ntimes n},$ is positive semi-definite if and only if it is of the form $X= xx^{T}$, for some $x in {-1,1 }^n$.
How can I prove this? Proving the 'if' part is easy since $y^Txx^Ty = (x^Ty)^T(x^Ty) geq 0$ for any $y in mathbb{R}^n$. But, the other way around is not that straightforward.
Notice that, $X$ has all ones diagonal since diagonal of a psd matrix cannot have negative elements.
linear-algebra matrices matrix-calculus symmetric-matrices positive-semidefinite
edited Dec 3 at 19:28
Yiorgos S. Smyrlis
62.3k1383162
asked Nov 22 at 18:45
independentvariable
13210
Claim.$,$ A matrix $,X in {-1,1}^{ntimes n},$ is positive semi-definite if and only if it is of the form $X= xx^{T}$, for some $x in {-1,1 }^n$.
How can I prove this? Proving the 'if' part is easy since $y^Txx^Ty = (x^Ty)^T(x^Ty) geq 0$ for any $y in mathbb{R}^n$. But, the other way around is not that straightforward.
Notice that, $X$ has all ones diagonal since diagonal of a psd matrix cannot have negative elements.
linear-algebra matrices matrix-calculus symmetric-matrices positive-semidefinite
linear-algebra matrices matrix-calculus symmetric-matrices positive-semidefinite
edited Dec 3 at 19:28
Yiorgos S. Smyrlis
62.3k1383162
asked Nov 22 at 18:45
independentvariable
13210
edited Dec 3 at 19:28
Yiorgos S. Smyrlis
62.3k1383162
asked Nov 22 at 18:45
independentvariable
13210
edited Dec 3 at 19:28
Yiorgos S. Smyrlis
62.3k1383162
edited Dec 3 at 19:28
Yiorgos S. Smyrlis
62.3k1383162
edited Dec 3 at 19:28
Yiorgos S. Smyrlis
62.3k1383162
62.3k1383162
asked Nov 22 at 18:45
independentvariable
13210
asked Nov 22 at 18:45
independentvariable
13210
asked Nov 22 at 18:45
independentvariable
13210
13210
add a comment |
add a comment |
1 Answer
1
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up vote
2
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accepted
Let $A$ be our matrix. The corresponding quadratic form should look like
$$
p(x_1,ldots,x_n)=boldsymbol x^tAboldsymbol x=x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j, tag{1}
$$
where $varepsilon=pm 1$ and $boldsymbol x=(x_1,ldots,x_n)$. We shall show that $p$ is positive semi-definite (psd), then
$$
p(x_1,ldots,x_n)=(varepsilon_1 x_1+cdots+varepsilon_n x_n)^2, tag{2}
$$
for suitable $varepsilon_j=pm 1$, $j=1,ldots,n$, in which case
$A=(varepsilon_1,ldots,varepsilon_n)(varepsilon_1,ldots,varepsilon_n)^t$.
If the expression $(1)$ is psd, then $$
varepsilon_{ij}varepsilon_{jk}=varepsilon_{ik},quadtext{for all $ine jne kne i$.}
tag{3}
$$
If $(3)$ fails, and for some $i,j,k$, we have
$varepsilon_{ij}varepsilon_{jk}=-varepsilon_{ik}$, then letting $boldsymbol xin mathbb R$ be the vector having $varepsilon_{jk},varepsilon_{ik},varepsilon_{ij}$ in the posistions $i,j,k$, respectively, and zero everywhere else, it can be readily seen that
$$
boldsymbol x^tAboldsymbol x=varepsilon_{jk}^2
+varepsilon_{ik}^2+varepsilon_{ij}^2+2
(varepsilon_{ij}varepsilon_{jk}+varepsilon_{ij}varepsilon_{ik}
+varepsilon_{jk}varepsilon_{ik})=3-6=-3<0.
$$
But if $(3)$ holds, then
$$
x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j=(x_1+varepsilon_{12}x_2+varepsilon_{12}varepsilon_{23}x_3+cdots+varepsilon_{12}cdotsvarepsilon_{n-1,n}x_n)^2
$$
Note. Another proof is based on the Pigeonhole Principle. There exist exactly $2^{n-1}$ symmetric matrices in ${-1,1}^{ntimes n},$ which satisfy $(3)$ and $varepsilon_{ii}=1$, for all $i$. Also, there exist exactly $2^{n-1}$ matrices of the form $boldsymbol xcdotboldsymbol x^t$, where $boldsymbol x$ contains elements in ${-1,1}$.
answered Nov 22 at 20:00
Yiorgos S. Smyrlis
62.3k1383162
1
How do you arrive at (3)? Why does a product of $epsilon$'s appear?
– daw
Nov 22 at 20:36
@daw To show that (3) is necessary for spd-ness, assume it does not hold, i.e., $varepsilon_{12}=-1$, $varepsilon_{23}=-1$, $varepsilon_{13}=-1$, and $boldsymbol x=(1,1,1)$, then $p(1,1,1)=-3$.
– Yiorgos S. Smyrlis
Nov 22 at 21:11
Can you please provide more detail? This is not very clear. The last part is not obvious.
– independentvariable
Nov 23 at 13:27
@independentvariable See my modified answer.
– Yiorgos S. Smyrlis
Nov 23 at 18:55
Ps: I believe (3) should hold for $i neq j neq k neq i$. Am I correct?
– independentvariable
Nov 23 at 18:56
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $A$ be our matrix. The corresponding quadratic form should look like
$$
p(x_1,ldots,x_n)=boldsymbol x^tAboldsymbol x=x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j, tag{1}
$$
where $varepsilon=pm 1$ and $boldsymbol x=(x_1,ldots,x_n)$. We shall show that $p$ is positive semi-definite (psd), then
$$
p(x_1,ldots,x_n)=(varepsilon_1 x_1+cdots+varepsilon_n x_n)^2, tag{2}
$$
for suitable $varepsilon_j=pm 1$, $j=1,ldots,n$, in which case
$A=(varepsilon_1,ldots,varepsilon_n)(varepsilon_1,ldots,varepsilon_n)^t$.
If the expression $(1)$ is psd, then $$
varepsilon_{ij}varepsilon_{jk}=varepsilon_{ik},quadtext{for all $ine jne kne i$.}
tag{3}
$$
If $(3)$ fails, and for some $i,j,k$, we have
$varepsilon_{ij}varepsilon_{jk}=-varepsilon_{ik}$, then letting $boldsymbol xin mathbb R$ be the vector having $varepsilon_{jk},varepsilon_{ik},varepsilon_{ij}$ in the posistions $i,j,k$, respectively, and zero everywhere else, it can be readily seen that
$$
boldsymbol x^tAboldsymbol x=varepsilon_{jk}^2
+varepsilon_{ik}^2+varepsilon_{ij}^2+2
(varepsilon_{ij}varepsilon_{jk}+varepsilon_{ij}varepsilon_{ik}
+varepsilon_{jk}varepsilon_{ik})=3-6=-3<0.
$$
But if $(3)$ holds, then
$$
x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j=(x_1+varepsilon_{12}x_2+varepsilon_{12}varepsilon_{23}x_3+cdots+varepsilon_{12}cdotsvarepsilon_{n-1,n}x_n)^2
$$
Note. Another proof is based on the Pigeonhole Principle. There exist exactly $2^{n-1}$ symmetric matrices in ${-1,1}^{ntimes n},$ which satisfy $(3)$ and $varepsilon_{ii}=1$, for all $i$. Also, there exist exactly $2^{n-1}$ matrices of the form $boldsymbol xcdotboldsymbol x^t$, where $boldsymbol x$ contains elements in ${-1,1}$.
answered Nov 22 at 20:00
Yiorgos S. Smyrlis
62.3k1383162
1
How do you arrive at (3)? Why does a product of $epsilon$'s appear?
– daw
Nov 22 at 20:36
@daw To show that (3) is necessary for spd-ness, assume it does not hold, i.e., $varepsilon_{12}=-1$, $varepsilon_{23}=-1$, $varepsilon_{13}=-1$, and $boldsymbol x=(1,1,1)$, then $p(1,1,1)=-3$.
– Yiorgos S. Smyrlis
Nov 22 at 21:11
Can you please provide more detail? This is not very clear. The last part is not obvious.
– independentvariable
Nov 23 at 13:27
@independentvariable See my modified answer.
– Yiorgos S. Smyrlis
Nov 23 at 18:55
Ps: I believe (3) should hold for $i neq j neq k neq i$. Am I correct?
– independentvariable
Nov 23 at 18:56
|
show 2 more comments
up vote
2
down vote
accepted
Let $A$ be our matrix. The corresponding quadratic form should look like
$$
p(x_1,ldots,x_n)=boldsymbol x^tAboldsymbol x=x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j, tag{1}
$$
where $varepsilon=pm 1$ and $boldsymbol x=(x_1,ldots,x_n)$. We shall show that $p$ is positive semi-definite (psd), then
$$
p(x_1,ldots,x_n)=(varepsilon_1 x_1+cdots+varepsilon_n x_n)^2, tag{2}
$$
for suitable $varepsilon_j=pm 1$, $j=1,ldots,n$, in which case
$A=(varepsilon_1,ldots,varepsilon_n)(varepsilon_1,ldots,varepsilon_n)^t$.
If the expression $(1)$ is psd, then $$
varepsilon_{ij}varepsilon_{jk}=varepsilon_{ik},quadtext{for all $ine jne kne i$.}
tag{3}
$$
If $(3)$ fails, and for some $i,j,k$, we have
$varepsilon_{ij}varepsilon_{jk}=-varepsilon_{ik}$, then letting $boldsymbol xin mathbb R$ be the vector having $varepsilon_{jk},varepsilon_{ik},varepsilon_{ij}$ in the posistions $i,j,k$, respectively, and zero everywhere else, it can be readily seen that
$$
boldsymbol x^tAboldsymbol x=varepsilon_{jk}^2
+varepsilon_{ik}^2+varepsilon_{ij}^2+2
(varepsilon_{ij}varepsilon_{jk}+varepsilon_{ij}varepsilon_{ik}
+varepsilon_{jk}varepsilon_{ik})=3-6=-3<0.
$$
But if $(3)$ holds, then
$$
x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j=(x_1+varepsilon_{12}x_2+varepsilon_{12}varepsilon_{23}x_3+cdots+varepsilon_{12}cdotsvarepsilon_{n-1,n}x_n)^2
$$
Note. Another proof is based on the Pigeonhole Principle. There exist exactly $2^{n-1}$ symmetric matrices in ${-1,1}^{ntimes n},$ which satisfy $(3)$ and $varepsilon_{ii}=1$, for all $i$. Also, there exist exactly $2^{n-1}$ matrices of the form $boldsymbol xcdotboldsymbol x^t$, where $boldsymbol x$ contains elements in ${-1,1}$.
answered Nov 22 at 20:00
Yiorgos S. Smyrlis
62.3k1383162
1
How do you arrive at (3)? Why does a product of $epsilon$'s appear?
– daw
Nov 22 at 20:36
@daw To show that (3) is necessary for spd-ness, assume it does not hold, i.e., $varepsilon_{12}=-1$, $varepsilon_{23}=-1$, $varepsilon_{13}=-1$, and $boldsymbol x=(1,1,1)$, then $p(1,1,1)=-3$.
– Yiorgos S. Smyrlis
Nov 22 at 21:11
Can you please provide more detail? This is not very clear. The last part is not obvious.
– independentvariable
Nov 23 at 13:27
@independentvariable See my modified answer.
– Yiorgos S. Smyrlis
Nov 23 at 18:55
Ps: I believe (3) should hold for $i neq j neq k neq i$. Am I correct?
– independentvariable
Nov 23 at 18:56
|
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $A$ be our matrix. The corresponding quadratic form should look like
$$
p(x_1,ldots,x_n)=boldsymbol x^tAboldsymbol x=x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j, tag{1}
$$
where $varepsilon=pm 1$ and $boldsymbol x=(x_1,ldots,x_n)$. We shall show that $p$ is positive semi-definite (psd), then
$$
p(x_1,ldots,x_n)=(varepsilon_1 x_1+cdots+varepsilon_n x_n)^2, tag{2}
$$
for suitable $varepsilon_j=pm 1$, $j=1,ldots,n$, in which case
$A=(varepsilon_1,ldots,varepsilon_n)(varepsilon_1,ldots,varepsilon_n)^t$.
If the expression $(1)$ is psd, then $$
varepsilon_{ij}varepsilon_{jk}=varepsilon_{ik},quadtext{for all $ine jne kne i$.}
tag{3}
$$
If $(3)$ fails, and for some $i,j,k$, we have
$varepsilon_{ij}varepsilon_{jk}=-varepsilon_{ik}$, then letting $boldsymbol xin mathbb R$ be the vector having $varepsilon_{jk},varepsilon_{ik},varepsilon_{ij}$ in the posistions $i,j,k$, respectively, and zero everywhere else, it can be readily seen that
$$
boldsymbol x^tAboldsymbol x=varepsilon_{jk}^2
+varepsilon_{ik}^2+varepsilon_{ij}^2+2
(varepsilon_{ij}varepsilon_{jk}+varepsilon_{ij}varepsilon_{ik}
+varepsilon_{jk}varepsilon_{ik})=3-6=-3<0.
$$
But if $(3)$ holds, then
$$
x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j=(x_1+varepsilon_{12}x_2+varepsilon_{12}varepsilon_{23}x_3+cdots+varepsilon_{12}cdotsvarepsilon_{n-1,n}x_n)^2
$$
Note. Another proof is based on the Pigeonhole Principle. There exist exactly $2^{n-1}$ symmetric matrices in ${-1,1}^{ntimes n},$ which satisfy $(3)$ and $varepsilon_{ii}=1$, for all $i$. Also, there exist exactly $2^{n-1}$ matrices of the form $boldsymbol xcdotboldsymbol x^t$, where $boldsymbol x$ contains elements in ${-1,1}$.
answered Nov 22 at 20:00
Yiorgos S. Smyrlis
62.3k1383162
Let $A$ be our matrix. The corresponding quadratic form should look like
$$
p(x_1,ldots,x_n)=boldsymbol x^tAboldsymbol x=x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j, tag{1}
$$
where $varepsilon=pm 1$ and $boldsymbol x=(x_1,ldots,x_n)$. We shall show that $p$ is positive semi-definite (psd), then
$$
p(x_1,ldots,x_n)=(varepsilon_1 x_1+cdots+varepsilon_n x_n)^2, tag{2}
$$
for suitable $varepsilon_j=pm 1$, $j=1,ldots,n$, in which case
$A=(varepsilon_1,ldots,varepsilon_n)(varepsilon_1,ldots,varepsilon_n)^t$.
If the expression $(1)$ is psd, then $$
varepsilon_{ij}varepsilon_{jk}=varepsilon_{ik},quadtext{for all $ine jne kne i$.}
tag{3}
$$
If $(3)$ fails, and for some $i,j,k$, we have
$varepsilon_{ij}varepsilon_{jk}=-varepsilon_{ik}$, then letting $boldsymbol xin mathbb R$ be the vector having $varepsilon_{jk},varepsilon_{ik},varepsilon_{ij}$ in the posistions $i,j,k$, respectively, and zero everywhere else, it can be readily seen that
$$
boldsymbol x^tAboldsymbol x=varepsilon_{jk}^2
+varepsilon_{ik}^2+varepsilon_{ij}^2+2
(varepsilon_{ij}varepsilon_{jk}+varepsilon_{ij}varepsilon_{ik}
+varepsilon_{jk}varepsilon_{ik})=3-6=-3<0.
$$
But if $(3)$ holds, then
$$
x_1^n+cdots+x_n^2+2sum_{1le i<jle n}varepsilon_{ij}x_ix_j=(x_1+varepsilon_{12}x_2+varepsilon_{12}varepsilon_{23}x_3+cdots+varepsilon_{12}cdotsvarepsilon_{n-1,n}x_n)^2
$$
Note. Another proof is based on the Pigeonhole Principle. There exist exactly $2^{n-1}$ symmetric matrices in ${-1,1}^{ntimes n},$ which satisfy $(3)$ and $varepsilon_{ii}=1$, for all $i$. Also, there exist exactly $2^{n-1}$ matrices of the form $boldsymbol xcdotboldsymbol x^t$, where $boldsymbol x$ contains elements in ${-1,1}$.
answered Nov 22 at 20:00
Yiorgos S. Smyrlis
62.3k1383162
edited Nov 25 at 15:32
answered Nov 22 at 20:00
Yiorgos S. Smyrlis
62.3k1383162
answered Nov 22 at 20:00
Yiorgos S. Smyrlis
62.3k1383162
answered Nov 22 at 20:00
Yiorgos S. Smyrlis
62.3k1383162
62.3k1383162
1
How do you arrive at (3)? Why does a product of $epsilon$'s appear?
– daw
Nov 22 at 20:36
@daw To show that (3) is necessary for spd-ness, assume it does not hold, i.e., $varepsilon_{12}=-1$, $varepsilon_{23}=-1$, $varepsilon_{13}=-1$, and $boldsymbol x=(1,1,1)$, then $p(1,1,1)=-3$.
– Yiorgos S. Smyrlis
Nov 22 at 21:11
Can you please provide more detail? This is not very clear. The last part is not obvious.
– independentvariable
Nov 23 at 13:27
@independentvariable See my modified answer.
– Yiorgos S. Smyrlis
Nov 23 at 18:55
Ps: I believe (3) should hold for $i neq j neq k neq i$. Am I correct?
– independentvariable
Nov 23 at 18:56
|
show 2 more comments
1
How do you arrive at (3)? Why does a product of $epsilon$'s appear?
– daw
Nov 22 at 20:36
@daw To show that (3) is necessary for spd-ness, assume it does not hold, i.e., $varepsilon_{12}=-1$, $varepsilon_{23}=-1$, $varepsilon_{13}=-1$, and $boldsymbol x=(1,1,1)$, then $p(1,1,1)=-3$.
– Yiorgos S. Smyrlis
Nov 22 at 21:11
Can you please provide more detail? This is not very clear. The last part is not obvious.
– independentvariable
Nov 23 at 13:27
@independentvariable See my modified answer.
– Yiorgos S. Smyrlis
Nov 23 at 18:55
Ps: I believe (3) should hold for $i neq j neq k neq i$. Am I correct?
– independentvariable
Nov 23 at 18:56
1
1
How do you arrive at (3)? Why does a product of $epsilon$'s appear?
– daw
Nov 22 at 20:36
How do you arrive at (3)? Why does a product of $epsilon$'s appear?
– daw
Nov 22 at 20:36
@daw To show that (3) is necessary for spd-ness, assume it does not hold, i.e., $varepsilon_{12}=-1$, $varepsilon_{23}=-1$, $varepsilon_{13}=-1$, and $boldsymbol x=(1,1,1)$, then $p(1,1,1)=-3$.
– Yiorgos S. Smyrlis
Nov 22 at 21:11
@daw To show that (3) is necessary for spd-ness, assume it does not hold, i.e., $varepsilon_{12}=-1$, $varepsilon_{23}=-1$, $varepsilon_{13}=-1$, and $boldsymbol x=(1,1,1)$, then $p(1,1,1)=-3$.
– Yiorgos S. Smyrlis
Nov 22 at 21:11
Can you please provide more detail? This is not very clear. The last part is not obvious.
– independentvariable
Nov 23 at 13:27
Can you please provide more detail? This is not very clear. The last part is not obvious.
– independentvariable
Nov 23 at 13:27
@independentvariable See my modified answer.
– Yiorgos S. Smyrlis
Nov 23 at 18:55
@independentvariable See my modified answer.
– Yiorgos S. Smyrlis
Nov 23 at 18:55
Ps: I believe (3) should hold for $i neq j neq k neq i$. Am I correct?
– independentvariable
Nov 23 at 18:56
Ps: I believe (3) should hold for $i neq j neq k neq i$. Am I correct?
– independentvariable
Nov 23 at 18:56
|
show 2 more comments
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