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Is it possible to design a square $d \times d$ matrix with

  1. value $2^n$ on the main diagonal ($n \geq 2$ not given)

  2. all off-diagonal elements chosen from the set $\{ 1, 2, 2^2, \dots, 2^{n-1} \}$

  3. not have full rank for any choice of $n$ or $d$

I think this is impossible (based on trial and error). But would appreciate a proof or a counterexample.

Rodrigo de Azevedo
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user11128
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  • Yes. Everything is positive actually – user11128 May 25 '20 at 13:48
  • What do you mean "based on trial and error"? – twosigma May 25 '20 at 13:49
  • Ok so $x=2^n$ ($n$ not given). And every other element is from the set ${1,2,4...,2^{n-1}}$. I created some 22 and 33 examples and it seems impossible based off that. I don't know how to argue it in general though. – user11128 May 25 '20 at 13:51
  • Yes I believe so – user11128 May 25 '20 at 13:58
  • Yes ideally I want to show that regardless of n and d, we always have full rank – user11128 May 25 '20 at 14:01
  • I haven't tried that – user11128 May 25 '20 at 14:07
  • At the very least, this is impossible for $n = 2$ (for all $d$) – Ben Grossmann May 25 '20 at 14:19
  • I asked what you meant by "trial and error" because I would be careful about coming to conclusions from just testing some examples (though it doesn't hurt to try), especially if you don't know much about the structure of your question. In general, if I pick random numbers to fill a matrix, it will almost certainly be full rank. Of course, your question here is not completely random, but the point is that it could be that the number of counterexamples is very small compared to the entirety of all possible cases. – twosigma May 25 '20 at 14:20
  • @twosigma But they did say "I think this is" rather than "I know this is" – Ben Grossmann May 25 '20 at 14:21
  • Yes, that's why I said "it doesn't hurt to try". I didn't mean to discourage him, I was just pointing out an important general remark in case he didn't already know that. – twosigma May 25 '20 at 14:22
  • @Omnomnomnom how do you know it is impossible for $n = 2$ and all $d$? Just curious since that could provide some insight. – twosigma May 25 '20 at 19:50
  • @twosigma yes I do. In particular, it is easy to see what the eigenvalues of the matrix are in this case since the matrix is a rank-1 perturbation of a multiple of the identity matrix. Alternatively, the sum of a positive definite matrix and a positive semidefinite matrix is necessarily positive definite. – Ben Grossmann May 25 '20 at 21:00
  • @Omnomnomnom Interesting. Thanks. – twosigma May 25 '20 at 21:27
  • @user11128 I think I can answer your original question, if you're still interested, but I'm not sure if it's appropriate to post an answer here because it's a different question. – twosigma May 25 '20 at 22:41

1 Answers1

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Some (hilariously inefficient) Matlab code to test this conjecture:

n = 3;
d = 4;
N = n^(d^2 - d);

for i = 0:N-1
    ch = dec2base(i,n,(d^2 - d));
    m = zeros(d^2 - d,1);
    for j = 1:(d^2 - d)
        m(j) = str2num(ch(j));
    end
    P = 2.^reshape(m,[d,d-1]);
    M = 2^n*eye(d);
    for j = 1:d
        col = 1:d;
        col(j) = [];
        M(j,col) = P(j,:);
    end
    if det(M) == 0
        disp('counterexample:')
        disp(M)
    end
end

There is indeed no counterexample for $n = 3,d=4$.

Ben Grossmann
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  • Could we argue that any system of eqns Mx=y with y=/=0 represents d equations in d unknowns. Since the maximum entry of K is on the diagonal, no one equation is a multiple of another and thus the system is consistent and thus there must exist a solution for x. By virtue of this, K must have been non-singular? – user11128 May 25 '20 at 20:41
  • @user11128 the fact that one equation is not a multiple of the other is not enough to conclude that the matrix is non-singular. For example, consider $$ \pmatrix{-2&1&1\1&-2&1\1&1&-2}. $$ – Ben Grossmann May 25 '20 at 20:57
  • Right, I get that we want no row being a linear combination of the others. Perhaps I'm going off on a tangent but I was wondering whether the fact that the maximum element is on the diagonal, might help simplify this somehow. – user11128 May 25 '20 at 21:01
  • @user1128 I see what you mean now. I don't see a clear way to make that argument work, though. – Ben Grossmann May 25 '20 at 21:04