How to prove that each element of $A(A^TA)^{-1}A^Ty$ is greater than 0, if $A(i,j)>0$ and $y=[1, 1, 1, ..., 1]^T$

Question

Let $A\in \mathbb{R}^{m\times n}$, $m>n$, $rank(A)=n$, and $\forall 1 \leq i \leq m, 1 \leq j \leq n, A(i, j)>0$, $y=[1, 1, 1, ..., 1]^T$. Let $\beta=A(A^TA)^{-1}A^Ty$, how to prove that each element of $\beta$ is positive?

Actually, I have tried a lot of random simulation experiments of this question and no one exception has occured, but I don't know how to prove it theoretically. enter image description here

Here is the code:

'''

count = 0;

while count<1000000

  count=count+1;
  col=randi([2,20]);
  A = 100*rand(10000,col);
  [m,n] = size(A);
  random_order = randperm(m);
  A = A(random_order, :);

  Y=ones(m,1);

  W = inv(A'*A)*A'*Y; 
  y = A*W;
  numNegatives = sum(y < 0);
  miny=min(y);

  if numNegatives>0
      break
      A
  end
  fprintf("%dth，%d rows，%d columns， %d Negatives\n",count,m, n, numNegatives)

end

'''

Results:

'''

999989th，10000 rows，12 columns， 0 Negatives
999990th，10000 rows，18 columns， 0 Negatives
999991th，10000 rows，4 columns， 0 Negatives
999992th，10000 rows，16 columns， 0 Negatives
999993th，10000 rows，12 columns， 0 Negatives
999994th，10000 rows，5 columns， 0 Negatives
999995th，10000 rows，14 columns， 0 Negatives
999996th，10000 rows，5 columns， 0 Negatives
999997th，10000 rows，2 columns， 0 Negatives
999998th，10000 rows，20 columns， 0 Negatives
999999th，10000 rows，16 columns， 0 Negatives
1000000th，10000 rows，19 columns， 0 Negatives

'''

Answer 1

This does not hold in general.

Note that $\beta=A(A^TA)^{-1}A^Ty\in \mathbb{R}^m$ belongs to the image ${\rm Im} A$ (that is, to the column space of $A$). Also, $A^T\beta=A^TA(A^TA)^{-1}A^Ty=A^Ty$, so $A^T(y-\beta)=0$, in other words, $y-\beta\in \operatorname{Ker} A^T=(\operatorname{Im} A)^\perp$. Thus, $\beta$ is just the projection of $y$ to $\operatorname{Im} A$. Can it have non-positive coordinates? Of course, why not. You may construct a counterexample already for $m=3$ and $n=2$. It is natural to start with choosing the projection of $y$ to $\operatorname{Im} A$, let this vector $v_0:=(a, b, c)^T$ be such that with $a<0,b>1$ and $a+b+c=a^2+b^2+c^2$ (that yields $v_0-y\perp v_0$). For example, you may choose $a=1/100,b=1+1/100$ and find appropriate $c$. Then take two columns of $A$ as two arbitrary linearly independent vectors with positive coordinates which are orthogonal to the vector $y-v_0=(1-a, 1-b,1-c)^T$. Since the vector $y-v_0$ have coordinates of both signs, you can find such vectors