For $n=10$ there exists such a matrix of rank(4): $$ \begin{pmatrix} \phantom{-}4 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & -3 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & -1 & -1 \\ \phantom{-}6 & \phantom{-}1 & \phantom{-}1 & \phantom{-}0 & \phantom{-}2 & \phantom{-}0 & -1 & -1 & \phantom{-}0 & \phantom{-}0 \\ \phantom{-}8 & \phantom{-}0 & \phantom{-}3 & -3 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & \phantom{-}3 & \phantom{-}1 & -2 \\ -3 & -1 & \phantom{-}0 & -1 & \phantom{-}0 & -1 & \phantom{-}0 & \phantom{-}1 & \phantom{-}0 & \phantom{-}0 \\ \phantom{-}0 & \phantom{-}0 & \phantom{-}1 & -1 & \phantom{-}2 & \phantom{-}0 & \phantom{-}0 & \phantom{-}1 & \phantom{-}1 & \phantom{-}0 \\ -7 & -1 & -1 & \phantom{-}0 & \phantom{-}1 & -1 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & \phantom{-}1 \\ -3 & \phantom{-}0 & -1 & \phantom{-}1 & -2 & \phantom{-}1 & \phantom{-}1 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0\\ \phantom{-}3 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & -1 & -1 & -1 & -1 & \phantom{-}0 \\ \phantom{-}0 & -1 & \phantom{-}0 & -1 & \phantom{-}0 & -2 & -1 & \phantom{-}0 & -1 & \phantom{-}0 \\ \phantom{-}0 & -2 & \phantom{-}0 & -2 & -9 & \phantom{-}0 & \phantom{-}2 & \phantom{-}4 & -1 & -3 \end{pmatrix} $$ Hence we can improve the upper bound to $c=\log_{10}(4)=0.6021$. I found this matrix using matlab. However I did not find any solution with $n=7$ and rank 3. To answer the question: Yes there exist solutions with only $-1,-0$ and $1$: $$ \begin{pmatrix} -1 & 0 & -1 & 0 & -1 & -1 & 0 & -1 & 0 & 0\\ 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1\\ 0 & -1 & 1 & -1 & 0 & 1 & 0 & 0 & 0 & -1\\ -1 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & -1 & 0\\ 0 & 0 & -1 & -1 & -1 & -1 & -1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & -1 & -1 & 1 & -1 & 1\\ 0 & 1 & -1 & 1 & 0 & 0 & 1 & -1 & 1 & 0\\ 0 & 0 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & 0\\ 1 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ -1 & 0 & 0 & 1 & 0 & 0 & 1 & -1 & 0 & -1 \end{pmatrix} $$ Here is the matlab code I used: function mathoverlow_triangular global n k nk n=10; k=4; nk=n*k; options=optimset('Jacobian','on');%'Display','iter',,'DerivativeCheck','on' exitflag=0; while exitflag~=1 %until a solution has been found [x,~,exitflag]=fsolve(@(x) f(x),rand(2*nk,1),options); end [~,~,A]=f(x) %find 0-1 pattern nz=double(abs(A)>=abs(A')); save('sol','nz'); for reps=1:1e4 %try this many times finding an integer solution p=randperm(n); ind=p(1:k); B1=nz; pm=1-2*(rand(n,k)>.5); is=1; B1(:,ind)=nz(:,ind).*pm;%.*randi(4,n,k) for j=k+1:n nu=null(B1(B1(:,p(j))==0,ind),'r'); %exclude those which give 0 on diagonal if size(nu,2)>0 nu=nu(:,B1(p(j),ind)*nu~=0); end if size(nu,2)==0 is=0; break; else B1(:,p(j))=B1(:,ind)*nu(:,randi(size(nu,2))); end end if is A=B1; [N,D]=rat(A); A=round(A.*(ones(n,1)*lcm_array(D))); A=round(A./(ones(n,1)*gcd_array(A))); A=round(A./(gcd_array(A')'*ones(1,n))); if max(abs(A(:)))==1 %write tex code for i=1:n stri=''; for j=1:n-1 stri=[stri num2str(A(i,j)) ' & ']; end disp([stri num2str(A(i,end)) '\\']); end disp(''); save('sol','A'); end end end end function [b]=lcm_array(A) if size(A,1)==1 b=A; else b=lcm(lcm_array(A(1:end-1,:)),A(end,:)); end end function [b]=gcd_array(A) if size(A,1)==1 b=A; else b=gcd(gcd_array(A(1:end-1,:)),A(end,:)); end end function [res,J,UV] = f(x) global n k nk U=reshape(x(1:nk),[n k]); V=reshape(x(nk+1:end),[n k]); UV=U*V'; Tnk=transposeT(n,k); res=UV'.*UV-eye(n); Tnn=transposeT(n,n); n2=n^2; dUV=sparse(1:n2,1:n2,UV(:),n2,n2); UVt=UV'; dUVt=sparse(1:n2,1:n2,UVt(:),n2,n2); J=(dUVt+dUV*Tnn)*[kron(V,speye(n)) kron(speye(n),U)*Tnk]; end function T = transposeT(n,k) %derivative of transpose map nk=n*k; u=reshape(1:nk,[n k]); v=u'; T=sparse(u(:),v(:),ones(nk,1),nk,nk); end