Decomposing a positive semi-definite matrix with all -1,+1 elements











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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







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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







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      up vote
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      up vote
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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







      share|cite|improve this question















      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






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      edited Dec 3 at 19:28









      Yiorgos S. Smyrlis

      62.3k1383162




      62.3k1383162










      asked Nov 22 at 18:45









      independentvariable

      11710




      11710






















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          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 ${-1,1}$.






          share|cite|improve this answer



















          • 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













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          1 Answer
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          1 Answer
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          up vote
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          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 ${-1,1}$.






          share|cite|improve this answer



















          • 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

















          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 ${-1,1}$.






          share|cite|improve this answer



















          • 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















          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 ${-1,1}$.






          share|cite|improve this answer














          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 ${-1,1}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 at 15:32

























          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
















          • 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




















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